The Number of Perfect Matchings in Hexagons on the Torus by Pfaffians

Let G be a (molecular) graph. A perfect matching of G is de ﬁ ned as a set of edges which are independent and cover every vertex of G exactly once. In the article, we present the formula on the number of the perfect matchings of two types of hexagons on the torus by Pfa ﬃ ans.


Introduction
A hexagonal system, or benzenoid system [1], or honeycomb lattice, is a finite connected subgraph of the infinite hexagonal lattice without cut vertices or non-hexagonal interior faces. Hansen and Zheng [2] and Gutman et al. [3] considered the problem about the normal components of benzenoid systems, respectively.
A perfect matching of a (molecular) graph G is defined as a set of edges which are independent and cover every vertex of G exactly once. It is also named as Kekulé structure in organic chemistry and closed-packed dimer in statistical physics, respectively. Denote the number of perfect matchings of G by ΦðGÞ. In the 1930s, Cyvin and Lovász et al. [4][5][6] firstly focused on the problems about the perfect matchings of a graph for two different and unrelated purposes. The number of perfect matchings or Kekulé structures had been widely used in various problems in the chemical fields [4,[7][8][9]. Cyvin et al. [10][11][12][13][14] discussed the enumeration of Kekulé structures for all kinds of benzenoids. Qian and Zhang presented the number of Kekulé structures for capped armchair nanotubes and capped zigzag nanotubes by transfer matrix method in [15,16], respectively. Yan et al. [17] further discussed the problem about the perfect matchings for one type of hexagons on a cylinder (cf. Figure 1(a)). Li and Zhang also obtained the number of per-fect matchings for two types of hexagons on the cylinder and the Möbius strip in [18] (cf. Figure 1).
Let G = ðVðGÞ, EðGÞÞ and H = ðVðHÞ, EðHÞÞ be two given graphs. Let G × H denote the Cartesian product of G with H, where VðG × HÞ = VðGÞ × VðHÞ, EðG × HÞ = ffðu 1 , u 2 Þ, ðv 1 , v 2 Þgju 1 = v 1 and fu 2 , v 2 g ∈ EðH Þ, or u 2 = v 2 and fu 1 , v 1 g∈ EðGÞg. A hexagonal cylinder, which is denoted by HS, of length 2m and breadth n is a graph obtained from the Cartesian product of a 2 m-cycle (x 1 u 1 x 2 u 2 ⋯ x m u m ) and an n-path (12 ⋯ n) by deleting the set of edges Let x 1 u 1 x 2 u 2 ⋯ x m u m and y 1 v 1 y 2 v 2 ⋯y m v m indicate the vertices of two cycles on the upper and lower boundaries of HS, respectively, where x i corresponds to y i and u i corresponds to v i ði = 12, ⋯, mÞ (cf. Figure 2). Without loss of generality, suppose further that both u i and v i are the vertices of degree 3. The graphs H 2m,n,r , 0 ≤ r ≤ bm/2c and i = 12, ⋯, m, are from HS by adding all edges x i y i−r , where i − r is modulo m (cf. Figure 2). Clearly, H 2m,n,r has a natural embedding on a torus.
The enumerations of perfect matchings of H 2m,n,r are considered by Wu [19] and Klein [20], where Klein gave the formula of the number of perfect matchings of H 2m,n,r by transfer matrix method when both n and r are odd or even. In the present article, we consider further the problem involving perfect matchings of H 2m,n,r by crossing orientation given by Tesler and the plane model of a graph.

Pfaffian Orientation
Let G = ðVðGÞ, EðGÞÞ be an undirected graph, where VðGÞ = f1, 2, 3,⋯,2pg is the vertex set of G. Let each edge fi, jg of a graph G have a weight denoted by ω ij . And the weight ω ij = 1 for all edges in each unweighted graph. Suppose that where Suppose that M=ffj 1 , j 1 ′ g,⋯,fj p , j p ′ gg is a perfect matching. Then the signed weight of the perfect matching M is where sgn 1 2 ⋯ 2p − 1 2p if the permutation is even ; One can define the Pfaffian of the matrix A as where M is over all perfect matchings of G.
For the perfect matching M, denote its signed weight by ω M . One can think the sign of ω M is just the sign of the perfect matchingM. For an orientation G ! of a graph, if the signs of its all perfect matchings are the same, then the orientation is a Pfaffian orientation of G. If a graph has a Pfaffian orientation, then the graph is named to be Pfaffian.
Kasteleyn [22] presented the Pfaffian orientations for planar graphs and also interpreted that the perfect matchings of a graph which is embedded on a surface with genus g can be computed as a linear combination of 4 g Pfaffians of the graph. Galllucio and Loebl [23], and Tesler [24] proved Kasteleyn's conclusion, independently. There are

Plane Model on H 2m,n,r
Let P denote a 4-polygon with four sides p 1 , p 2 , p 1 ′ , p 2 ′. Suppose that G is a graph embedded on a torus. A drawing of a graph G is defined as a plane model of G, in which if all the edges of the graph G could be divided into E 0 , E 1 and E 2 , and not only the subgraph induced by E 0 is a spanning plane graph, which wholly contained inside P, but also the edges in E 1 and E 2 from the sides p j to p j ′ of P do not cross.
Noting that H 2m,n,r is a graph which may be embedded on a torus, now we can give a plane model of H 2m,n,r by the definition above-mentioned such that the edges of H 2m,n,r are separated into E 0 , E 1 and E 2 . Moreover, the subgraph induced by E 0 is a spanning plane graph. The spanning plane graph is wholly contained inside P, and its edges in E 1 and E 2 from the sides p j and p j ′ of P do not cross (cf. Figure 2). If n is odd, a plane model of H 2m,n,r is shown as in Figure 3(b). If n is even, a plane model of H 2m,n,r is shown as in Figure 4(b). It is obvious that every edge in E 1 crosses each every in E 2 exactly once. For simplicity, suppose that 1, 2, ⋯, 2mn are 2mn vertices of H 2m,n,r and EðH 2m,n,r Þ denote the edge set. Then E 0 , E 1 and E 2 are represented as the followings: (1) If n is even, then we have (2) If n is even, then we have An orientation of a graph embedded on a torus in a plane model is the crossing orientation if it conforms to the rule of cross orientation in reference [24]. For the graph G which is embedded on torus, let Xðx 1 , x 2 Þ be its skew adjacent matrix, with the edges in E 0 having weight 1 and the edges in E j having weight x j ðj = 1, 2Þ. A formula computing the number for perfect matchings in G was presented by Tesler [24]:

Theorem 3 ([24]).
(a) A graph may be oriented such that every perfect matching M has sign where the constant ω 0 = ±1 could be explained as the sign of a perfect matching without crossing edges if the edges exist, and κðMÞ denotes the number of times edges in M cross.

(b) An orientation of a graph satisfies (a) if and only if the orientation is a crossing orientation
According to the crossing orientation by Tesler, one crossing orientation of H 2m,n,r where n is even or n is odd is indicated in Figures 5 and 6. Figure 5(a) and Figure 5(b) give an orientations of all the edges E 0 ∪ E 1 and E 0 ∪ E 2 when n is even, respectively. Figures 6(a) and 6(b) give the orientations of the edges E 0 ∪ E 1 and E 0 ∪ E 2 when n is odd, respectively.

The Sign Weights of Perfect
Matchings of H 2m,n,r Let the weight of edge of E 0 in H 2m,n,r be 1. Suppose that x and y are the weight of edge of E 1 (Referring to Figures 5(a) and 6(a)) and E 2 (Referring to Figures 5(b) and 6(b)) in H 2m,n,r , respectively. Let Xðx, yÞ ðx, y = ±1Þ be the skew adjacency matrix of H 2m,n,r if n is even and Yðx, y Þ ðx, y = ±1Þ the skew adjacency matrix of H 2m,n,r if n is odd. To determine the sign of Pfaffians Pf Xðx, yÞ and Pf Y ðx, yÞ, the perfect matchings in H 2m,n,r are distinguished into four classes. The perfect matchings belonging to class 1 are those that have odd number of edges both in E 1 and in E 2 . The perfect matchings belonging to class 2. have odd number of edges in E 1 and even number of edges in E 2 . The perfect matchings belonging to class 3 have even number of edges in E 1 and odd number of edges in E 2 . The perfect matchings belonging to class 4 have even number of edges both in E 1 and in E 2 . Clearly, κðMÞ is always even, where the perfect matching M lies in three classes other than class 1.
It is convenient to consider the case when y = ±1. For simplicity, let x and x 2 denote the odd power and even power, respectively. In light of the method deciding the sign of perfect matchings by Lu, Zhang and Lin [29], we also can obtain the sings of all the perfect matchings in H 2m,n,r . If y = 1, the signs of perfect matchings in three classes other than class 1 are always positive by Theorem 3. If y = −1, the signs of perfect matchings in three classes other than class 3 are always positive by Theorem 3 because that the perfect matching belonging to class 1 contains an odd of number edges in E 2 and κðMÞ is always even. Therefore, we can decide the sign weights of all the perfect matchings of H 2m,n,r shown as in Table 1.
From Table 1, we can observe that the signs of perfect matchings belonging to classes 3 and 4 when y = 1 are always positive and the signs of perfect matchings in classes 1 and 2 when y = 1 are always positive. Then by formula (11) we have the following result.

Classs
The signs of the perfect matchings x, y = 1 x, y = −1

Journal of Mathematics
Proof. Recall that H ! m,2n,r is a crossing orientation in the plane model of H 2m,n,r . Then the elements of XðH ! m,2n,r Þ which can be read off from Figure 5 have the following form when n is even: where X ij ðx, yÞ is the n × n matrix. If i ≤ j, If i > j, then X ij = −X T ji (X T ji is the transpose of X ij ). Let A i ði = 1, 2, 3, 4Þ be an n × n matrix as follows: Then the matrices AðxÞ, B, C, BðyÞ and CðyÞ can be expressed by the following forms: Therefore by formula (15) we obtain that Journal of Mathematics Now we turn to calculate the determinant det ðJ t Þ of J t . By formula (28), we have where and It is convenient to calculate that we adjust appropriately the order of labeling for the vertices of H 2m,n,r such that the determinant det ðJ t Þ turn out to be the following: Let φ t = ð2t + 1Þπ/m and θ t = 2tπ/m. Then it is not difficult to obtain the expressions of ω t , ω −1 t and ω t + ω −1 t when y = 1 and y = −1 by formula (29). Thus we have the determinant det ðJ t Þ if y = 1, i.e., If y = −1, then we have det J t ð Þ = −1 ð Þ n/2 2 n cos θ t − 1 ð Þ n − 2 n/2 −1 ð Þ n/2 + 1 Â Ã cos θ t − 1 ð Þ n/2 −1 ð Þ n/2 cos rθ t + i sin rθ t ð Þ + cos rθ t − i sin rθ t ð Þ Â Ã x + x 2 :