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Further results on enumeration of perfect matchings of Cartesian product graphs

  • Tingzeng Wu EMAIL logo and Xiaolin Zeng
Published/Copyright: September 24, 2024

Abstract

Counting perfect matchings is an interesting and challenging combinatorial task. It has important applications in statistical physics and chemistry. As the general problem is #P-complete, it is usually tackled by randomized heuristics and approximation schemes. Let G and H be two graphs. Denote by G × H the Cartesian product of graphs G and H . The number of perfect matchings of some types of Cartesian product G × H is investigated by the Pfaffian method, where G P 2 , P 3 , P 4 or C 4 , and H T or H is a non-bipartite graph with a unique cycle. In this article, we construct a Pfaffian orientation of the graph G × P 2 , where G is a non-bipartite graph with k odd cycles. We obtain the counting formula on the number of perfect matchings of G × P 2 .

MSC 2010: 05A15; 05C30; 05C50; 05C70

1 Introduction

Let G = ( V ( G ) , E ( G ) ) be a simple connected graph with vertex set V ( G ) = { v 1 , v 2 , , v n } and edge set E ( G ) = { e 1 , e 2 , , e m } . The Cartesian product of two graphs G and H is defined as

V ( G × H ) = V ( G ) × V ( H ) ,

where x = u 1 v 1 and y = u 2 v 2 , u 1 , u 2 V ( G ) , v 1 , v 2 V ( H ) , then

E ( G × H ) = { ( x , y ) u 1 = u 2 and v 1 v 2 E ( H ) , or v 1 = v 2 and u 1 u 2 E ( G ) } .

Suppose that G is an arbitrary orientation of G , and C is an arbitrary cycle with even length. We call C is oddly oriented in G if C contains an odd number of edges which are directed in G that the direction of each orientation of C . A cycle C is said to be nice if G C has a perfect matching, where G C means the induced subgraph of G obtained from G by removing the vertices of C and all incident edges. If every nice cycle of even length of G is oddly oriented in G , we say that G is a Pfaffian orientation of G . A graph G is said to be Pfaffian if it admits a Pfaffian orientation.

Let G be a simple graph, and let G = ( V ( G ) , E ( G ) ) be an arbitrary orientation of G . The skew adjacent matrix A ( G ) = [ a i j ] of G is defined as

a i j = 1 if  ( v i , v j ) E ( G ) , 1 if  ( v j , v i ) E ( G ) , 0 otherwise .

A matching in a graph is a set of isolated edges with no shared endpoints. And a perfect matching in a graph is a matching that saturates every vertex. The number of perfect matchings of G is denoted by M ( G ) . The cycle C is a nice cycle in G if C is a nice cycle in a spanning subgraph of G . An F-alternating cycle of G is a cycle whose edges are alternately in E ( G ) F if F is a perfect matching of G . The path on n vertices is denoted by P n .

The matching problem on a graph is equivalent to a physical model of dimers. This was mostly studied on planar graphs (lattices), where there is a Pfaffian method by Kasteleyn [1], which shows how to exactly count dimer arrangements (perfect matchings). The number of perfect matchings is an important topological index that has been applied for the estimation of the resonant energy and total π -electron energy and calculation of pauling bond order [2,3]. Counting the number of perfect matchings of a graph is #P-complete problem even in the bipartite graph. Hence, many physicists, mathematicians, and chemists have focused most of their attention on the enumeration of perfect matchings of graphs [413].

Kasteleyn first gave a polynomial time algorithm for enumeration of the perfect matchings in planar graphs using the Pfaffian method and extended his approach to toroidal grids in [1,14]. Little [15] generalized Kasteleyn’s work and proved that if a bipartite graph G has no subdivision of K 3 , 3 , then G has a Pfaffian orientation. In recent years, Yan and Zhang [10,16] Yan and Yeh [11] achieved some significant results in the aspect of studying Pfaffian orientation. They proved that if G is a graph on n vertices and there exists no subgraph which is an even subdivision of K 2 , 3 after the contraction of at most one cycle of even length, then G has a Pfaffian orientation and presented an expression formula to enumerate the number of perfect matchings of a reflective symmetric graph. In particular, they have given the computing formula of enumerating the number of perfect matchings of some types of Cartesian product graphs G × H by the Pfaffian method, where G P 2 , P 3 , P 4 , or C 4 , and H T . Lu and Zhang [17] characterized the Pfaffian property of Cartesian products G × P 2 n and G × C 2 n for any graph G in terms of forbidden subgraphs of G . This extended the results obtained by Yan and Zhang [11]. Lin and Zhang [18] first considered the non-bipartite graphs. They proved that if G is a non-bipartite graph with a unique cycle, then G × P 2 has a Pfaffian orientation, and they characterized the explicit formula to enumerate the number of perfect matchings of G × P 2 . In this article, our interest is to investigate the Pfaffian property of Cartesian product G × P 2 , where G is a non-bipartite graph with k odd cycles.

The rest of this article is organized as follows. In Section 2, we present some notions and lemmas. In Section 3, we show that if G is a cycle-chain graph, then G × P 2 is Pfaffian. We characterize the computing formula to enumerate perfect matchings of G × P 2 . In the final section, we prove that if G is a petaling graph, then G × P 2 is Pfaffian. We also give the computing formula to enumerate perfect matchings of G × P 2 .

2 Preliminaries

Lemma 2.1

[19] Let G be a simple graph with an even number of vertices, and G be an orientation of G. Then, the following three properties are equivalent:

  1. G is a Pfaffian orientation of G.

  2. Every nice cycle of even length of G is oddly oriented in G .

  3. If G contains a perfect matching, then for some perfect matching F, every F-alternating cycle is oddly oriented in G .

Yan and Zhang [11] introduced an oriented method for Cartesian product graph G × P 2 . The method is described as follows: let G be a simple graph, and let G be an orientation of G . We take two copies of G and get two graphs G 1 and G 2 with V ( G 1 ) = { v 1 , v 2 , , v n } , V ( G 2 ) = { v 1 , v 2 , , v n } , and add edges v i v i ( i = 1 , 2 , , n ) between every pair of corresponding vertices v i V ( G 1 ) and v i V ( G 2 ) , respectively. Then, the resulting graph is G × P 2 . The method of orienting the resulting graph is given as follows: G 1 is an orientation of G 1 , the orientation of G 2 is reversing the orientation of each of edges in G 1 , and the directions of edges v i v i ( i = 1 , 2 , , n ) are from v i to v i . Then, we obtain the oriented graph G × P 2 of G × P 2 .

Lemma 2.2

[11] Suppose that T is a tree with V ( T ) = v 1 , v 2 , , v n . Then, every cycle C of T × P 2 is a nice cycle and has the following form:

v i 1 v i 2 v i m 1 v i m v i m v i m 1 v i 2 v i 1 v i 1 .

where i 1 , i 2 , , i m { 1 , 2 , , n } .

Lemma 2.3

[19,20] Let G be a graph on n vertices. If G is a Pfaffian orientation of G, then

[ M ( G ) ] 2 = det A ( G ) ,

where A ( G ) is the skew adjacent matrix of G .

3 Pfaffian orientation and enumeration of perfect matchings for the Cartesian product of k odd unicyclic chain graph and P 2

To begin with, we present some notions. A graph is called unicyclic if it contains exactly one cycle, and the size of edges is equal to its vertices. For a unicyclic graph, if the induced cycle is odd, then the unicyclic graph is called odd unicyclic graph. Suppose that T i and U k denote a tree and an odd unicyclic graph, respectively. The graph G is called a k odd unicyclic chain graph obtained by joining a vertex of T i and a vertex of the induced cycle of odd unicyclic graph U i , and joining a vertex of T i and a vertex of induced cycle of odd unicyclic graph U i + 1 , where i = 1 , 2 , , k 1 . The graph is depicted in Figure 1. We can see that the vertices of T i that join vertices of the induced cycle of U i and U i + 1 can be the same one, and the vertices of the induced cycle of U j that join vertices of T j 1 and T j are different, where j = 2 , 3 , , k 1 .

Figure 1 
               The 
                     
                        
                        
                           k
                        
                        k
                     
                   odd unicyclic chain graph.
Figure 1

The k odd unicyclic chain graph.

Take two copies of G , denoted by G and G . Assume that the odd unicyclic graphs of G and G are U 1 , U 2 , , U k and U 1 , U 2 , , U k resp., V ( G ) = { v 1 , v 2 , , v n } and V ( G ) = { v 1 , v 2 , , v n } . By adding edges v i v i ( i = 1 , 2 , , n ) , the resulting graph is the Cartesian product of k odd unicyclic chain graph and P 2 .

Theorem 3.1

Let G be a k odd unicyclic chain graph (Figure 1). Then, G × P 2 is Pfaffian.

Proof

We prove this theorem by induction.

If k = 2 , it means that G is a bicyclic graph and non-bipartite. We mark the vertices of induced cycle in U 1 as v 1 , v 2 , , v 2 j + 1 , and mark the vertices of induced cycle in U 2 as v s , v s + 1 , , v t , mark the vertices of induced cycle in U 1 as v 1 , v 2 , , v 2 j + 1 , and mark the vertices of induced cycle in U 2 as v s , v s + 1 , , v t .

Suppose that E = { v 1 v 2 j + 1 , v s v t , v 1 v 2 j + 1 , v s v t } , and C is any even cycle of G × P 2 . We agree that G v i ( i = 1 , 2 , ) is obtained by deleting edges v i from G and edges v i from G . Then, we discuss the cases of C as follows.

Case 1. If C contains no edge of E , it implies that C is an even cycle in ( G v 1 v 2 j + 1 v s v t ) × P 2 that is a spanning subgraph of G × P 2 , where G v 1 v 2 j + 1 v s v t is obtained by deleting edges v 1 v 2 j + 1 and v s v t from G , and edges v 1 v 2 j + 1 and v s v t from G . Since G v 1 v 2 j + 1 v s v t is a tree, by Lemma 2.2, we know that C is a nice cycle. Obviously, the cycle C is also a nice cycle of G × P 2 . By Lemma 2.1, we derive that G × P 2 is a Pfaffian orientation of G × P 2 .

Case 2. If C contains an edge of E . Assume that C contains edge v 1 v 2 j + 1 . Since G v 1 v 2 j + 1 v s v t is a tree, that is to say it is a bipartite graph. Suppose that ( V 1 , V 2 ) is a bipartition of it, so the bipartition of ( G v 1 v 2 j + 1 v s v t ) × P 2 is ( V 1 V 2 , V 1 V 2 ) , where V i and V i are corresponding to V i ( i = 1 , 2 ) . Moreover, there exists a path with even length v 2 j + 1 v x v x + 1 v x + 1 v x v 1 , so we can deduce that the vertices v 1 and v 2 j + 1 belong to the same partition. Hence, C v 1 v 2 j + 1 is a path P v 1 v 2 j with even length of ( G v 1 v 2 j + 1 v s v t ) × P 2 , i.e., C is an odd cycle, which is a contradiction. Analogously, if C contains edge v s v t or v 1 v 2 j + 1 or v s v t , C is also an odd cycle.

Case 3. If C contains two edges of E .

Subcase 1. Suppose that C contains edges v 1 v 2 j + 1 and v 1 v 2 j + 1 . Then, there exist two disjoint paths P v 1 v 1 and P v 2 j + 1 v 2 j + 1 of ( G v 1 v 2 j + 1 v s v t ) × P 2 if C E . From Lemma 2.2, we obtain that P v 1 v 1 = v 1 v 1 or v 1 v i 1 v i a v i a v i 1 v 1 , P v 2 j + 1 v 2 j + 1 = v 2 j + 1 v 2 j + 1 or v 2 j + 1 v j 1 v j b v j b v j 1 v 2 j + 1 . Therefore, { v r v r 1 < r n , v r , v r V ( C ) } is a perfect matching of ( G × P 2 ) C . So C is a nice cycle. Since the form of C is v 1 v 2 j + 1 v i m 1 v i m v i m v i m 1 v 2 j + 1 v 1 v 1 , v i m v i m 1 v s v t , i m 1 , i m 1 , 2 , , n , from the definition of G × P 2 , we obtain that the orientation of edges v 1 v 2 j + 1 , , v i m 1 v i m are reversing the orientation of edges v 1 v 2 j + 1 , , v i m 1 v i m . Hence, C is oddly oriented. Similarly, if C contains edges v s v t and v s v t , then C is also oddly oriented. By Lemma 2.1, we obtain that G × P 2 is a Pfaffian orientation of G × P 2 .

Subcase 2. Assume that C contains the edges v 1 v 2 j + 1 and v s v t . Then, we obtain that there always exist two paths of even length so that it is asymmetrical and disjoint, which means that each of them contains odd vertices. So, there exists no perfect matching of ( G × P 2 ) C , that is to say that C is not a nice cycle in G × P 2 , which is a contradiction. Analogously, if C contains edges v 1 v 2 j + 1 and v s v t , there is also no perfect matching of G × P 2 , i.e., C is not a nice cycle in G × P 2 .

Case 4. If C contains three edges of E . Generally, suppose that C contains v 1 v 2 j + 1 , v 1 v 2 j + 1 , and v s v t . Similar to the consideration of Case 2, C v 1 v 2 j + 1 is a path, i.e., it is a bipartite graph. Since there exists a path of even length P v 1 v 2 j + 1 = v 2 j + 1 v i 1 v i 1 + 1 v s v t v i 1 + 1 v i 1 v 2 j + 1 v 1 v 1 , the vertices v 1 and v 2 j + 1 belong to the same partition. Therefore, C v 1 v 2 j + 1 is an even path P v 1 v 2 j + 1 of ( G v 1 v 2 j + 1 v s v t ) × P 2 , then we conclude that C has odd length, which is a contradiction. Similarly, if C contains v 1 v 2 j + 1 , v s v t , and v s v t , or v 1 v 2 j + 1 , v 1 v 2 j + 1 , and v s v t , or v s v t , v 1 v 2 j + 1 , and v s v t , we obtain that C is also an odd cycle of G × P 2 .

Case 5. If C contains all edges of E . With a similar consideration of Subcase 1 of Case 3, there exist four disjoint paths P v 1 v 1 , P v 2 j + 1 v s , P v t v t , and P v s v 2 j + 1 if C E . By Lemma 2.2, we have P v 1 v 1 = v 1 v 1 or v 1 v c 1 v c a v c a v c 1 v 1 , P v 2 j + 1 v s = v 2 j + 1 v d 1 v d b v i , P v t v t = v t v t or v t v e a v e b v e b v e a v t , P v s v 2 j + 1 = v i v d b v d 1 v 2 j + 1 . So { v r v r 1 < r n , v r , v r V ( C ) } is a perfect matching of ( G × P 2 ) C , and then C is a nice cycle. Since the form of C is v 1 v 2 j + 1 v s v t v i m v i m v t v s v 2 j + 1 v 1 v 1 , i m { 1 , 2 , , n } , by the definition of G × P 2 , C is oddly oriented in G × P 2 . Hence, by Lemma 2.1, the G × P 2 is a Pfaffian orientation of G × P 2 .

So the theorem holds for k = 2 , and we can see that if C is an even cycle and oddly oriented, C must contain corresponding edges of E , that is to say if C contains edge v i v j , then it must contain v i v j .

Now, assume that the theorem holds for k = w .

When k = w + 1 , let E be an edge set that selects an edge from the induced cycle of U 1 , U 2 , , U w in G , and selects a corresponding edge from the induced cycle of U 1 , U 2 , , U w in G . E is an edge set that chooses an edge from the induced cycle of U w + 1 in G and a corresponding edge from the induced cycle of U w + 1 in G . Then, we discuss the cases of C as follows that C is any even cycle of G × P 2 . Since the theorem holds for k = w , it implies that C contains no edge or contain even corresponding edges of E .

Case 1. If C contains no edge of E .

Subcase 1. Suppose that C also contains no edge of E , then C is an even cycle in ( G v 1 v 2 j + 1 v s v t ) × P 2 that is a spanning subgraph of G × P 2 by removing the edges of E and E . Since G × P 2 is a tree after deleting all edges from E and E , then C is a nice cycle of this spanning subgraph. Apparently, the cycle C is also a nice cycle of G × P 2 . By Lemma 2.1, we derive that G × P 2 is a Pfaffian orientation of G × P 2 .

Subcase 2. If C also contains any edge of E , by the arguments as above, we obtain that the length of C is always odd, which is a contradiction.

Subcase 3. Assume that C also contains two edges of E , then C has even length and they are corresponding edges. So, there always exist even and disjoint paths, which means that there exists a perfect matching of ( G × P 2 ) C . Hence, C is a nice cycle. From the definition of G × P 2 , we have C is oddly oriented in G × P 2 . Therefore, by Lemma 2.1, G × P 2 is a Pfaffian orientation of G × P 2 .

Case 2. If C contains even corresponding edges of E .

Subcase 1. If C also contains no edge of E . Since the theorem holds for k = w , obviously, G × P 2 is a Pfaffian orientation of G × P 2 .

Subcase 2. Suppose that C also contains an edge of E , from the arguments above, we know that the C always has odd length, which is a contradiction.

Subcase 3. Assume that C also contains two edges of E , then C still has even corresponding edges. So, we obtain that there exist even and disjoint paths if C E E . By Lemma 2.2, we have ( G × P 2 ) C contains a perfect matching, so C is a nice cycle. By the definition of G × P 2 , we know that C is oddly oriented in G × P 2 . By Lemma 2.1, we derive that G × P 2 is a Pfaffian orientation of G × P 2 .

Therefore, the theorem holds for k = w + 1 . The theorem is proved by induction.□

Theorem 3.2

Let G be a k-odd unicyclic chain graph with n vertices. Then

M ( G × P 2 ) = j = 1 n 1 α j 2 ,

where the product ranges over all eigenvalues α j of A ( G ) .

Proof

By Theorem 3.1, we can find an orientation G × P 2 such that it is a Pfaffian orientation of G × P 2 . With an appropriate ordering of the vertices of G × P 2 , we have the skew adjacent matrix of G × P 2 is

A ( G × P 2 ) = A ( G ) I n I n A ( G ) ,

where I n is the n × n unit matrix. By Lemma 2.3, we obtain that

[ M ( G × P 2 ) ] 2 = det A ( G × P 2 ) = det A ( G ) I n I n A ( G ) = det ( I n A ( G ) 2 ) .

Since A ( G ) is a real skew adjacent matrix, the eigenvalues of it are 0 or pure imaginary, so we have

det ( I n A ( G ) 2 ) = j = 1 n ( 1 α j 2 ) ,

where α j is the eigenvalue of A ( G ) . Therefore, we derive that

M ( G × P 2 ) = j = 1 n 1 α j 2 .

Figure 2 
               
                  
                     
                        
                        
                           G
                           ×
                           
                              
                                 P
                              
                              
                                 2
                              
                           
                        
                        G\times {P}_{2}
                     
                   and its oriented graph.
Figure 2

G × P 2 and its oriented graph.

Example 1

Enumerate the number of perfect matchings of G × P 2 (Figure 2). Checking Figure 2, it is easy to obtain that the skew adjacent matrix of G is

A ( G ) = 0 1 1 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 .

So, the eigenvalues of A ( G ) are 0, ± i , ± i 5 , ± i 2 ( 6 + 2 ) , and ± i 2 ( 6 2 ) . By Theorem 3.2, we obtain that M ( G × P 2 ) = 72 .

4 Pfaffian orientation and enumeration of perfect matchings for the Cartesian product of k odd petaling graph and P 2

Let U 1 , U 2 , , U k be k odd unicyclic graphs, and let C i be the induced cycle of U i , where i = 1 , 2 , , k . The graph G is called k odd unicyclic petaling graph obtained by identifying a vertex of cycle C 1 in U 1 , cycle C 2 in U 2 , , cycle C k in U k .The graph is depicted in Figure 3.

Figure 3 
               The 
                     
                        
                        
                           k
                        
                        k
                     
                   odd unicyclic petaling graph.
Figure 3

The k odd unicyclic petaling graph.

Similarly, we take two copies of G and obtain two graphs G with odd unicyclic graphs U 1 , U 2 , , U k , and G with odd unicyclic graphs U 1 , U 2 , , U k , V ( G ) = { v 1 , v 2 , , v n } , V ( G ) = { v 1 , v 2 , , v n } , and add edges v i v i ( i = 1 , 2 , , n ) between every pair of corresponding vertices v i V ( G ) and v i V ( G ) , respectively. The resulting graph is the Cartesian product of k odd unicyclic petaling graph and P 2 .

Theorem 4.1

Let G be a k odd unicyclic petaling graph. Then, G × P 2 is Pfaffian.

Proof

Similar to the proof of Theorem 3.1, we prove this theorem by induction.

If k = 3 , it means that G is a tricyclic graph and non-bipartite. The graph G is obtained by identifying a vertex of cycle C 1 in U 1 , cycle C 2 in U 2 , and cycle C 3 in U 3 , and we label the intersected vertex as v i . We label the vertices of C 1 in U 1 as v 1 , v 2 , , v i , , v 2 k + 1 , label the vertices of C 2 in U 2 as v i , , v j , v j + 1 , label the vertices of C 3 in U 3 as v i , , v s , v s + 1 , label the vertices of C 1 in U 1 as v 1 , v 2 , , v i , , v 2 k + 1 , label the vertices of C 2 in U 2 as v i , , v j , v j + 1 , and label the vertices of C 3 in U 3 as v i , , v s , v s + 1 .

Assume that E = { v 1 v 2 k + 1 , v j v j + 1 , v s v s + 1 , v 1 v 2 k + 1 , v j v j + 1 , v s v s + 1 } , and C is an even cycle of G × P 2 . We agree that G v i ( i = 1 , 2 , ) is obtained by deleting edges v i from G and edges v i from G . In the following, we consider the cases of C .

Case 1. If C contains no edge of E , it means that C is an even cycle in ( G v 1 v 2 k + 1 v j v j + 1 v s v s + 1 ) × P 2 that is a spanning subgraph of G × P 2 , where G v 1 v 2 k + 1 v j v j + 1 v s v s + 1 means that the graph is obtained by deleting edges v 1 v 2 k + 1 , v j v j + 1 , and v s v s + 1 from G , and edges v 1 v 2 k + 1 , v j v j + 1 , and v s v s + 1 from G . Since G v 1 v 2 k + 1 v j v j + 1 v s v s + 1 is a tree, by Lemma 2.2, we obtain that C is a nice cycle of ( G v 1 v 2 k + 1 v j v j + 1 v s v s + 1 ) × P 2 . Obviously, the cycle C is also a nice cycle of G × P 2 . By Lemma 2.1, G × P 2 is a Pfaffian orientation of G × P 2 .

Case 2. If C contains an edge of E . Suppose that C contains edge v 1 v 2 k + 1 . Since G v 1 v 2 k + 1 v j v j + 1 v s v s + 1 is a tree, i.e., it is a bipartite graph. Assume that ( V 1 , V 2 ) is a bipartition of it, so the bipartition of ( G v 1 v 2 k + 1 v j v j + 1 v s v s + 1 ) × P 2 is ( V 1 V 2 , V 1 V 2 ) , where V i and V i are corresponding to V i ( i = 1 , 2 ) . Moreover, there has a path with even length v 2 k + 1 v x v x + 1 v x + 1 v x v 1 , so we can deduce that the vertices v 1 and v 2 k + 1 belong to the same partition. Hence, C v 1 v 2 k + 1 is a path P v 1 v 2 k with even length of ( G v 1 v 2 k + 1 v j v j + 1 v s v s + 1 ) × P 2 , which implies that C is an odd cycle, which is a contradiction. Analogously, if C contains edge v 1 v 2 k + 1 or v j v j + 1 or v j v j + 1 or v s v s + 1 or v s v s + 1 , C is also an odd cycle.

Case 3. If C contains two edges of E .

Subcase 1. Suppose that C contains edges v 1 v 2 k + 1 and v 1 v 2 k + 1 . Then, there exist two disjoint paths P v 1 v 1 and P v 2 k + 1 v 2 k + 1 of ( G v 1 v 2 k + 1 v j v j + 1 v s v s + 1 ) × P 2 if C v 1 v 2 k + 1 v 1 v 2 k + 1 . By Lemma 2.2, we obtain that P v 1 v 1 = v 1 v 1 or v 1 v i 1 v i a v i a v i 1 v 1 , P v 2 k + 1 v 2 k + 1 = v 2 k + 1 v 2 k + 1 or v 2 k + 1 v j 1 v j b v j b v j 1 v 2 k + 1 . Hence, { v r v r 1 < r n , v r , v r V ( C ) } is a perfect matching of ( G × P 2 ) C . So C is a nice cycle. Since the form of C is v 1 v 2 k + 1 v i m 1 v i m v i m v i m 1 v 2 k + 1 v 1 v 1 , i m 1 , i m 1 , 2 , , n . From the definition of G × P 2 , we obtain that the orientation of edges v 1 v 2 k + 1 , , v i m 1 v i m is reversing the orientation of edges v 1 v 2 k + 1 , , v i m 1 v i m . Hence, C is oddly oriented. Similarly, if C contains edges v j v j + 1 and v j v j + 1 , or v s v s + 1 and v s v s + 1 , the C is also oddly oriented. By Lemma 2.1, we obtain that G × P 2 is a Pfaffian orientation of G × P 2 .

Subcase 2. Suppose that C contains edges v 1 v 2 k + 1 and v j v j + 1 . Then, we know that there always has two even paths that are asymmetrical and disjoint if ( G × P 2 ) C , which implies that each of them contains odd vertices. Hence, it has no perfect matching of ( G × P 2 ) C , i.e., C is not a nice cycle in G × P 2 , which is a contradiction. Analogously, if C contains edges v 1 v 2 k + 1 and v s v s + 1 or v j v j + 1 or v s v s + 1 , v j v j + 1 and v 1 v 2 k + 1 or v s v s + 1 , v s v s + 1 and v 1 v 2 k + 1 or v j v j + 1 , there also exists no perfect matching of ( G × P 2 ) C , i.e., C is not a nice cycle.

Case 4. If C contains three edges of E . Assume that C contains v 1 v 2 k + 1 , v j v j + 1 , and v 1 v 2 k + 1 . Similar to the consideration of Case 2, C v 1 v 2 k + 1 is a path, i.e., it is a bipartite graph. Since there exists a path with even length P v 1 v 2 k + 1 = v 2 k + 1 v i 1 v i 1 + 1 v j v j + 1 v i 1 + 1 v i 1 v 2 k + 1 v 1 v 1 , the vertices v 1 and v 2 k + 1 belong to the same partition. Therefore, C v 1 v 2 k + 1 is an even path P v 1 v 2 k + 1 , then we conclude that C has odd length, which is a contradiction. Analogously, if C contains v 1 v 2 k + 1 , v j v j + 1 , and v j v j + 1 , or v 1 v 2 k + 1 , v s v s + 1 , and v 1 v 2 k + 1 , or v 1 v 2 k + 1 , v s v s + 1 , and v s v s + 1 , or v j v j + 1 , v s v s + 1 , and v j v j + 1 , or v j v j + 1 , v s v s + 1 , and v s v s + 1 , or v 1 v 2 k + 1 , v 1 v 2 k + 1 , and v j v j + 1 , or v j v j + 1 , v 1 v 2 k + 1 , and v j v j + 1 , or v 1 v 2 k + 1 , v 1 v 2 k + 1 , and v s v s + 1 , or v s v s + 1 , v 1 v 2 k + 1 and v s v s + 1 , or v j v j + 1 , v j v j + 1 and v s v s + 1 , or v s v s + 1 , v j v j + 1 , and v s v s + 1 , we obtain that C is also an odd cycle of G × P 2 .

Case 5. If C contains four edges of E .

Subcase 1. Suppose that C contains v 1 v 2 k + 1 , v j v j + 1 , v 1 v 2 k + 1 and v j v j + 1 . With a similar consideration of Subcase 1 of Case 3, there exist four disjoint paths P v 1 v 1 , P v 2 k + 1 v j + 1 , P v j v j + 1 , and P v j + 1 v 2 k + 1 if C v 1 v 2 k + 1 v j v j + 1 v 1 v 2 k + 1 v j v j + 1 . By Lemma 2.2, we know that P v 1 v 1 = v 1 v 1 or v 1 v c 1 v c m v c m v c 1 v 1 , P v 2 k + 1 v j + 1 = v 2 k + 1 v d 1 v d l v j + 1 , P v j v j = v j v j or v j v e 1 v e u v e u v e 1 v j , P v j + 1 v 2 k + 1 = v j + 1 v d l v d 1 v 2 k + 1 . Therefore, we obtain that { v r v r 1 < r n , v r , v r V ( C ) } is a perfect matching of G C , and then C is a nice cycle. Since the form of C is v 1 v 2 k + 1 v j v j + 1 v i l 1 v i l v i l v i l l v j + 1 v j v 2 k + 1 v 1 v 1 , i l 1 , i l { 1 , 2 , , n } . By the definition of G × P 2 , C is oddly oriented in G × P 2 . Hence, by Lemma 2.1, the G × P 2 is a Pfaffian orientation of G × P 2 . Similarly, if C contains edges v 1 v 2 k + 1 , v s v s + 1 , v 1 v 2 k + 1 , and v s v s + 1 , or v j v j + 1 , v s v s + 1 , v j v j + 1 , and v s v s + 1 , the C is oddly oriented. In addition, G × P 2 is a Pfaffian orientation of G × P 2 .

Subcase 2. Assume that C contains four edges of E but they are not corresponding. With a similar proof of Subcase 2 of Case 3, we derive that it has no perfect matching of ( G × P 2 ) C , i.e., C is not a nice cycle in G × P 2 , which is a contradiction.

Case 6. If C contains five edges of E . Similar to the proof of Cases 2 and 4, we obtain that C is an odd cycle of G × P 2 , which is a contradiction.

Case 7. If C contains all edges of E . With a similar consideration of Case 5, we obtain that G × P 2 is a Pfaffian orientation of G × P 2 .

Therefore, we obtain that the theorem holds for k = 2 , and we know that if C is a nice cycle and oddly oriented, C must contain corresponding edges of E .

Suppose that the theorem holds for k = w .

If k = w + 1 , let E be an edge set that selects an edge from the induced cycle of U 1 , U 2 , , U w in G , respectively, and selects an edge from the induced cycle of U 1 , U 2 , , U w in G , respectively. In addition, E is an edge set that chooses an edge from the induced cycle of U w + 1 in G and a corresponding edge from the induced cycle of U w + 1 in G . Because the theorem holds for k = w , it implies that C contains no edges of E , or contains even edges and these edges are corresponding.

Case 1. If C contains no edge of E .

Subcase 1. Suppose that C also contains no edge of E , then C is an even cycle that is a spanning subgraph of G × P 2 by removing the edges E and E . Because G × P 2 is a tree after deleting all edges from E and E , then C is a nice cycle of this spanning subgraph. By Lemma 2.1, we know that G × P 2 is a Pfaffian orientation of G × P 2 .

Subcase 2. If C also contains any an edge of E , we obtain that C is always oddly, which is a contradiction.

Subcase 3. Assume that C also contains all edges of E , then C has even length and they are corresponding edges. From the argument above, we derive that G × P 2 is a Pfaffian orientation of G × P 2 .

Case 2. If C contains even edges of E and they are correspondence.

Subcase 1. If C also contains no edge of E . Since the theorem holds for k = w , apparently, G × P 2 is a Pfaffian orientation of G × P 2 .

Subcase 2. Suppose that C also contains an edge of E , from the arguments above, we obtain that the C always has odd length, which is a contradiction.

Subcase 3. Assume that C also contains two edges of E , C still has even corresponding edges. Hence, we obtain that there exist even and disjoint paths if C E E . By Lemma 2.2, we obtain that ( G × P 2 ) C contains a perfect matching, i.e., C is a nice cycle. From the definition of G × P 2 , we know that C is oddly oriented in G × P 2 . By Lemma 2.1, we know that G × P 2 is a Pfaffian orientation of G × P 2 .

Therefore, the theorem also holds for k = w + 1 . The theorem is proved.□

Theorem 4.2

Let G be a k odd unicyclic petaling graph with n vertices. Then

M ( G × P 2 ) = j = 1 n 1 α j 2 ,

where the product ranges over all eigenvalues α j of A ( G ) .

Proof

By Theorem 4.1, we can find an orientation such that G × P 2 is a Pfaffian orientation of G × P 2 . With a similar proof of Theorem 3.2, we can derive the result of this theorem.□

Figure 4 
               
                  
                     
                        
                        
                           G
                           ×
                           
                              
                                 P
                              
                              
                                 2
                              
                           
                        
                        G\times {P}_{2}
                     
                   and its oriented graph.
Figure 4

G × P 2 and its oriented graph.

Example 2

Enumerate the number of perfect matchings of G × P 2 (Figure 4).

From Figure 4, we obtain that the eigenvalues of A ( G ) are 0 of multiplicity 3, ± i 2 of multiplicity 3, ± i 2 ( 14 + 6 ) and ± i 2 ( 14 6 ) . By Theorem 4.2, we have M ( G × P 2 ) = 405 .

5 Conclusion

In this article, we prove that the two types of Cartesian product graphs are Pfaffians. We give the computing formulas on enumeration of perfect matchings of two Cartesian product graphs. Using the same method as above, we can show that some Cartesian product graphs are Pfaffians, and we can also obtain the counting formulas of enumerating perfect matchings of these graphs, such as the Cartesian product of G × P 2 , where G is a graph obtained by attaching some cycles to some leaves of a tree.

Acknowledgements

The authors are grateful for the reviewer’s valuable comments that improved the manuscript.

  1. Funding information: This research was supported by the National Natural Science Foundation of China (Nos 12261071 and 11761056) and the Natural Science Foundation of Qinghai Province (2020-ZJ-920).

  2. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results, and approved the final version of the manuscript. XZ provided all figures. TW prepared the manuscript with contributions from all co-authors.

  3. Conflict of interest: The authors declare that they have no conflict of interest.

  4. Data availability statement: No data were used to support this study.

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Received: 2022-11-04
Revised: 2024-08-02
Accepted: 2024-08-16
Published Online: 2024-09-24

© 2024 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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