Home Mathematics On the number of perfect matchings in random polygonal chains
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On the number of perfect matchings in random polygonal chains

  • Shouliu Wei EMAIL logo , Yongde Feng , Xiaoling Ke and Jianwu Huang
Published/Copyright: December 6, 2023

Abstract

Let G be a graph. A perfect matching of G is a regular spanning subgraph of degree one. Enumeration of perfect matchings of a (molecule) graph is interest in chemistry, physics, and mathematics. But the enumeration problem of perfect matchings for general graphs (even in bipartite graphs) is non-deterministic polynomial (NP)-hard. Xiao et al. [C. Xiao, H. Chen, L. Liu, Perfect matchings in random pentagonal chains, J. Math. Chem. 55 (2017), 1878–1886] have studied the problem of perfect matchings for random odd-polygonal chain (i.e., with odd polygons). In this article, we further present simple counting formulae for the expected value of the number of perfect matchings in random even-polygonal chains (i.e., with even polygons). Based on these formulae, we obtain the average values of the number for perfect matchings with respect to the set of all even-polygonal chains with n polygons.

MSC 2010: 05C05; 05C30

1 Introduction

Let G = ( V , E ) be a graph, where V ( G ) and E ( G ) are the vertex set and the edge set of G , respectively. A perfect matching or a 1-factor of G is a regular spanning subgraph of degree one. In other words, a perfect matching of G is a set of independent edges in E ( G ) covering every vertex exactly once. The perfect matching is also called Kekulé structure in organic chemistry and closed-packed dimer in statistical physics. Denote the number of perfect matchings of G by M ( G ) . All notations not defined in this study can be found in the book [1].

The enumeration problems for maximum matchings and perfect matchings of a graph play an important role in graph theory and combinatorial optimization and have a wide applications in some field. In organic chemistry, there are strong connections between the number of the Kekulé structures and chemical properties for benzenoid hydrocarbons, coronoid hydrocarbons, and nonbenzenoid hydrocarbons such as azulenoid kekulene [2]. For instance, those edges that are present in comparatively few of the Kekulé structures of a (molecule) graph turn out to correspond to the bonds that are least stable, and the more Kekulé structures a (molecule) graph possesses, the more stable the corresponding benzenoid molecule will be [3]. Additionally, the number of Kekulé structures is an important topological index, which had been applied estimating the resonant energy and total π -electron energy [2,4], calculating the Pauling bond order [5] and Clar aromatic sextet [6]. In crystal physics, the perfect matching problem is closely related to the dimer problem [79]. But the enumeration problem for perfect matchings in general graphs (even in bipartite graphs) is non-deterministic polynomial (NP)-hard [10,11]. Hence, it makes sense to seek special classes for which the problem can be solved exactly [10]. For a survey of results and further bibliography on the perfect matching, see [2,10,12,13] and reference cited therein.

It is well known that the fusions of hydrocarbons composed of polygonal rings can be represented as a natural graph called polygonal systems [14], where the vertices symbolize carbons. Precisely, a polygonal system is a connected graph without cut-vertices in which all inner faces are regular polygons (or cycles) of length one, such that any two polygons are either disjoint or have exactly one common edge, and no three polygons share a common edge. For example, the benzenoid hydrocarbons correspond to hexagonal systems, which were profoundly studied by many researchers [2,1519]. An important class of fused polygonal rings is the linear fusions of hydrocarbons composed of polygonal rings (or named catafusense polycyclic conjugated compounds, see [20]). Some results on this class were reported by Balaban [16,20], which include the enumeration of fusions in such class and the isomers in the fusions. The corresponding graph of this class is the so-called polygonal chain.

The characteristic graph of a given polygonal system G consists of vertices corresponding to polygons of G and edges, where there is an edge connecting two vertices if and only if the corresponding polygons share an edge in G . For example, a hexagonal system and its characteristic graph are shown in Figure 1. A polygonal system is called a polygonal chain if its characteristic graph is isomorphic to a simple path. For example, all the corresponding graphs of chamaazulene, phenylethylamine, linear fusion of cyclopentane module, unbranched catacondensed benzenoid, unbranched catacondensed polyomino, indenoinene, and unbranched catacondensed cyclooctane are polygonal chain, as shown in Figure 2. Polygonal chains have attracted chemists, physicists, and mathematicians for more than half century [2,7,8]. A polygonal chain with n polygons is called an r-polygonal chain if all of its polygons (or cycles) are r -polygons and denoted by P C n r . Suppose that each polygon (or cycle) of P C n r is labeled by C 1 r , C 2 r , , C n r . The polygon C i r will be called the ith polygon of P C n r , 1 i n . Note that there are many ways to fuse the two consecutive cycles. Thus, P C n r is not unique.

Figure 1 
               A hexagonal system and its characteristic graph.
Figure 1

A hexagonal system and its characteristic graph.

Figure 2 
               (a) Chamaazulene, (b) phenylethylamine, (c) linear fusion of cyclopentance modules, (d) unbranched catacondensed benzenoid, (e) unbranched catacondensed polyomino, (f) indenoinene, and (g) unbranched catacondensed cyclooctane.
Figure 2

(a) Chamaazulene, (b) phenylethylamine, (c) linear fusion of cyclopentance modules, (d) unbranched catacondensed benzenoid, (e) unbranched catacondensed polyomino, (f) indenoinene, and (g) unbranched catacondensed cyclooctane.

For a polygonal chain P C n r , if r is odd, then it is called odd-polygonal chain. Otherwise, it is called even-polygonal chain. Xiao et al. [2123] have considered the problem of the number for the perfect matchings in an explicit odd-polygonal chain or even-polygonal chain. And in the remark part of [22,23], they mentioned that one can use some operations, any polygonal chain R except the polygonal chain with odd number of odd polygons, and there exists a caterpillar tree T such that the Hosoya index of T equals the number of perfect matchings of R . In other words, they presented a method to enumerate the number for the perfect matchings of any polygonal chain except the polygonal chain with odd number of odd polygons. Actually, there may not be perfect matchings for the polygonal chain with odd number of odd polygons. Here, we deal with the problem on enumerating the number of random even-polygonal chains. For convenience, assume that r = 2 k in the rest of this article. Bearing in mind that P C n r is an even-polygonal chain with n r -polygons labeled by C 1 r , C 2 r , , C n r so that C i r and C i + 1 r are adjacent for i , where n 3 and 1 i n 1 . Both the first polygon C 1 r and the last polygon C n r are called terminal polygon. And the remaining polygons C 2 r , C 3 r , , C n 1 r are called internal polygon. By symmetry, each internal polygon is of type-1, type-2, ,type-(K-2), or type-(K-1) according to whether it separates its two adjacent polygons by a distance of 1,2, , k 2 , or k 1 from left to right, as shown in Figure 3, respectively.

Figure 3 
               
                  k − 1 types of internal polygons.
Figure 3

k − 1 types of internal polygons.

A random even-polygonal chain of length n is an even-polygonal chain with n polygons in which each internal polygon is one of type- i with the probability p i for 1 i k 1 , denoted by P C n r ( p 1 , p 2 , , p k 1 ) , where i = 1 k 1 p i = 1 . We assume that the probabilities p 1 , p 2 , , p k 1 are constants and independent of n , i.e., the process described above is a zeroth-order Markov process. Gutman [24,25] studied the problem of perfect matchings about random benzenoid chain graphs in the 1990s. Chen and Zhang [26] obtained an explicit analytical expression for the expected value of the number of perfect matchings in a random phenylene chain. A simple formula for the expected value of the number of perfect matchings in random polyomino chain graphs was presented in [12] and [27], respectively. Wei and Shiu [28] obtained the similar results about the number of perfect matchings and its asymptotic behavior in random polyazulenoid chains. A simple explicit expression for the expected value of the number of perfect matchings in a random odd-polygonal chain was presented in [21]. Recently, Wei et al. [29] also obtained the same results for random octagonal chain graphs.

In this study, simple explicit formulae are further obtained for the expected value of the number of perfect matchings in random even-polygonal chains and for the asymptotic behavior of this expectation. Based on the formulae, we also give the average value of the number of perfect matchings with respect to the set of all even-polygonal chains with n polygons.

2 Recursive formula for the number of perfect matchings

Recall the recursive formula for the perfect matchings in a graph G [2], i.e.,

(1) M ( G ) = M ( G u v ) + M ( G e ) ,

where e = u v denotes an edge of G incident with the vertices u and v .

Lemma 2.1

Let P C n r be an even-polygonal chain with n polygons, where r = 2 k . Then,

  1. For n = 1 , 2 ,

    M ( P C 1 r ) = 2 a n d M ( P C 2 r ) = 3 .

  2. For n 3 and if the ( n 1 ) -st polygon is of type-1, type-3, , type- α , then

    M ( P C n r ) = M ( P C n 1 r ) + M ( P C n 2 r ) ,

    where α = k 2 when k is odd, and α = k 1 when k is even.

  3. For n 3 and if the ( n 1 ) -st polygon is of type-2, type-4, , type- β , then

    M ( P C n r ) = 2 M ( P C n 1 r ) M ( P C n 2 r ) ,

    where β = k 1 when k is odd, and β = k 2 when k is even.

Proof

For n = 1 , 2 , it is obvious that M ( P C 1 r ) = 2 and M ( P C 2 r ) = 3 .

For n 3 , without loss of generality, let e = u v be an edge in the even-polygonal chain P C n r , as shown in Figures 4 to 7. We prove Conclusions 2 and 3 according to the type of the ( n 1 ) -st polygon.

Case 1. Suppose that the ( n 1 ) -st polygon is of type-1 shown in Figures 4 and 5. There are two types of perfect matchings in P C n r , where one of them includes the edge e = u v and the other does not include e . If k is odd, there is a unique perfect matching with the edge e = u v (denoted by the double edges, as shown in Figure 4) for the graph that consists of the n -st and ( n 1 ) -st polygons. Thus,

M ( P C n r u v ) = M ( P C n 2 r ) .

There is also a unique perfect matching without the edge e = u v (denoted by the double edges, as shown in Figure 4) for the n -st polygon. Thus,

M ( P C n r e ) = M ( P C n 1 r ) .

If k is even, there is a unique perfect matching with the edge e = u v (denoted by the double edges, as shown in Figure 5) for the n -st polygon. Thus,

M ( P C n r u v ) = M ( P C n 1 r ) .

There is also a unique perfect matching without the edge e = u v (denoted by the double edges, as shown in Figure 5) for the graph that consists of the n -st and ( n 1 ) -st polygons. Thus,

M ( P C n r e ) = M ( P C n 2 r ) .

Therefore, by Formula (1), we have the conclusion, i.e.,

M ( P C n r ) = M ( P C n 1 r ) + M ( P C n 2 r ) .

For the cases that the ( n 1 ) -st polygon is of type-1, type-3, , type- α , all the proofs are similar to those in Case 1, where α = k 2 when k is odd, and α = k 1 when k is even. Therefore, we have the result, i.e.,

M ( C C n ) = M ( C C n 1 ) + M ( C C n 2 ) .

Case 2. Suppose that the ( n 1 )st polygon is of type-2 shown in Figures 6 and 7. Being similar to the illustrations in Case 1, one can also obtain the results. If k is odd, it is easy to see from Figure 6 that

M ( P C n r u v ) = M ( P C n 1 r ) M ( P C n 1 r e ) = M ( P C n 1 r ) M ( P C n 2 r )

and

M ( P C n r e ) = M ( P C n 1 r ) .

If k is even, it is easy to see from Figure 7 that

M ( P C n r u v ) = M ( P C n 1 r )

and

M ( P C n r e ) = M ( P C n 1 r ) M ( P C n 1 r e ) = M ( P C n 1 r ) M ( P C n 2 r ) .

Therefore, by Formula (1), we have the conclusion, i.e.,

M ( P C n r ) = 2 M ( P C n 1 r ) M ( P C n 2 r ) .

For the other cases that the ( n 1 ) -st polygon is of type-2, type-4, , type- β , all the proofs are same to those in case 2, where β = k 1 when k is odd, and β = k 2 when k is even. Thus, we obtain the conclusion, i.e.,

M ( P C n r ) = 2 M ( P C n 1 r ) M ( P C n 2 r ) .

This completes the proof.□

Figure 4 
               Illustrations of type-1 when 
                     
                        
                        
                           k
                        
                        k
                     
                   is odd.
Figure 4

Illustrations of type-1 when k is odd.

Figure 5 
               Illustrations of type-1 when 
                     
                        
                        
                           k
                        
                        k
                     
                   is even.
Figure 5

Illustrations of type-1 when k is even.

Figure 6 
               Illustrations of type-2 when 
                     
                        
                        
                           l
                        
                        l
                     
                   is odd.
Figure 6

Illustrations of type-2 when l is odd.

Figure 7 
               Illustrations of type-2 when 
                     
                        
                        
                           l
                        
                        l
                     
                   is even.
Figure 7

Illustrations of type-2 when l is even.

3 Expected value of the number of perfect matchings

Note that the probabilities p 1 , p 2 , , p k 2 and p k 1 are unknown constants. Here, M ( P C n r ( p 1 , p 2 , , p k 1 ) ) is a random variable. Let E [ M ( P C n r ( p 1 , p 2 , , p k 1 ) ) ] be the expected value of M ( P C n r ( p 1 , p 2 , , p k 1 ) ) .

Lemma 3.1

Let P C n r ( p 1 , p 2 , , p k 1 ) be a random even-polygonal chain with n polygons, where n 3 and r = 2 k .

  1. If k is even, then

    E [ M ( P C n r ( p 1 , p 2 , , p k 1 ) ) ] = 1 + i = 1 k 2 2 p 2 i E [ M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) ] + 1 2 i = 1 k 2 2 p 2 i E [ M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ] .

  2. If k is odd, then

    E [ M ( P C n r ( p 1 , p 2 , , p k 1 ) ) ] = 1 + i = 1 k 1 2 p 2 i E [ M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) ] + 1 2 i = 1 k 1 2 p 2 i E [ M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ] .

Proof

Recall that the ( n 1 ) -st polygon of P C n r ( p 1 , p 2 , , p k 1 ) is one of type- i with probabilities p i for 1 i k 1 . If k is even, by Lemma 2.1, we have

E [ M ( P C n r ( p 1 , p 2 , , p k 1 ) ) ] = p 1 [ M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) + M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ] + p 2 [ 2 M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ] + p 3 [ M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) + M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ] + p 4 [ 2 M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ] + p k 2 [ 2 M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ] + p k 1 [ M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) + M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ] = [ ( p 1 + p 3 + + p k 1 ) + 2 ( p 2 + p 4 + + p k 2 ) ] M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) + [ ( p 1 + p 3 + + p k 1 ) ( p 2 + p 4 + + p k 2 ) ] M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) = 1 + i = 1 k 2 2 p 2 i M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) + 1 2 i = 1 k 2 2 p 2 i M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) .

If k is odd, by Lemma 2.1, we have

E [ M ( P C n r ( p 1 , p 2 , , p k 1 ) ) ] = p 1 [ M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) + M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ] + p 2 [ 2 M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ] + p 3 [ M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) + M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ] + p 4 [ 2 M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ] + p k 2 [ M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) + M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ] + p k 1 [ 2 M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ] = [ ( p 1 + p 3 + + p k 2 ) + 2 ( p 2 + p 4 + + p k 1 ) ] M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) + [ ( p 1 + p 3 + + p k 2 ) ( p 2 + p 4 + + p k 1 ) ] M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) = 1 + i = 1 k 1 2 p 2 i M ( C C ( n 1 ; p 1 , p 2 , , p k 1 ) ) + 1 2 i = 1 k 1 2 p 2 i M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) .

The last equality follows from p 1 + p 2 + + p k 1 = 1 .

Note that E [ E [ M ( P C n r ( p 1 , p 2 , , p k 1 ) ) ] ] = E [ M ( P C n r ( p 1 , p 2 , , p k 1 ) ) ] . Since E [ M ( P C n r ( p 1 , p 2 , , p k 1 ) ) ] is a sum of random variables, it is true that

E [ M ( P C n r ( p 1 , p 2 , , p k 1 ) ) ] = 1 + i = 1 k 2 2 p 2 i E [ M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) ] + 1 2 i = 1 k 2 2 p 2 i E [ M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ] .

when k is even, and

E [ M ( P C n r ( p 1 , p 2 , , p k 1 ) ) ] = 1 + i = 1 k 1 2 p 2 i E [ M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) ] + 1 2 i = 1 k 1 2 p 2 i E [ M ( P C n 2 r ( p 1 , p 2 , , p k 1 ) ) ]

when k is odd.□

Theorem 3.2

Let P C n r ( p 1 , p 2 , , p k 1 ) be a random even-polygonal chain with n polygons for n 1 .

  1. If k is even and i = 1 k 2 2 p 2 i < 1 (or k is odd and i = 1 k 1 2 p 2 i < 1 ), then

    (2) E [ M ( P C n r ( p 1 , p 2 , , p k 1 ) ) ] = 2 s 3 s t t n 1 2 t 3 s t s n 1 ,

    where s = 1 + i = 1 k 2 2 p 2 i 3 i = 1 k 2 2 p 2 i 2 4 2 and t = 1 + i = 1 k 2 2 p 2 i + 3 i = 1 k 2 2 p 2 i 2 4 2 when k is even, and s = 1 + i = 1 k 1 2 p 2 i 3 i = 1 k 1 2 p 2 i 2 4 2 and t = 1 + i = 1 k 1 2 p 2 i + 3 i = 1 k 1 2 p 2 i 2 4 2 when k is odd.

  2. If k is even and i = 1 k 2 2 p 2 i = 1 (or k is odd and i = 1 k 1 2 p 2 i = 1 ), then

    E [ M ( P C n r ( p 1 , p 2 , , p k 1 ) ) ] = 1 + n .

Proof

Let y n = E [ M ( P C n + 1 r ( p 1 , p 2 , , p k 1 ) ) ] with n 0 . Since M ( P C 1 r ) = 2 and M ( P C 2 r ) = 3 , we have the initial conditions y 0 = 2 and y 1 = 3 .

If k is even, by the first conclusion in Lemma 3.1, it follows that for n 0 ,

(3) y n + 2 = 1 + i = 1 k 2 2 p 2 i y n + 1 + 1 2 i = 1 k 2 2 p 2 i y n .

Note that the characteristic equation of Formula (3) is

(4) λ 2 1 + i = 1 k 2 2 p 2 i λ 1 2 i = 1 k 2 2 p 2 i = 0 .

Therefore, the characteristic roots of Formula (4) are

s = 1 + i = 1 k 2 2 p 2 i 1 + i = 1 k 2 2 p 2 i 2 + 4 1 2 i = 1 k 2 2 p 2 i 2 = 1 + i = 1 k 2 2 p 2 i 3 i = 1 k 2 2 p 2 i 2 4 2

and

t = 1 + i = 1 k 2 2 p 2 i + 1 + i = 1 k 2 2 p 2 i 2 + 4 1 2 i = 1 k 2 2 p 2 i 2 = 1 + i = 1 k 2 2 p 2 i + 3 i = 1 k 2 2 p 2 i 2 4 2 .

Case 1. If i = 1 k 2 2 p 2 i < 1 , then the two roots are distinct. In this case,

y n = k 1 s n + k 2 t n .

Substituting the initial conditions y 0 = 2 and y 1 = 3 , we obtain

k 1 + k 2 = 2 , k 1 s + k 2 t = 3 .

The solution of the equations above is

k 1 = 2 t 3 s t and k 2 = 2 s 3 s t .

Therefore,

y n = E [ M ( P C n + 1 r ( p 1 , p 2 , , p k 1 ) ) ] = 2 t 3 s t s n + 2 s 3 s t t n ,

which proves the first statement of the theorem.

Case 2. If i = 1 k 2 2 p 2 i = 1 , then s = t = 1 . In this case,

y n = k 1 + k 2 n .

Substituting the initial conditions y 0 = 2 and y 1 = 3 , we have

k 1 = 2 and k 2 = 1 ,

which means that y n = E [ M ( P C n + 1 r ( p 1 , p 2 , , p k 1 ) ) ] = 2 + n .

For the case when k is odd, we can prove the conclusion by the same discussion.□

By Theorem 3.2, we can obtain the expected value of the number of perfect matchings in random 6-polygonal chain (i.e., hexagonal chain). The random 6-polygonal chain P C n 6 ( p 1 , p 2 ) of length n is a random hexagonal chain graph with n hexagons [24,25] and we will denote it by H n ( p 1 , p 2 ) , where p 1 + p 2 = 1 .

Corollary 3.3

[24,25] Let H n ( p 1 , p 2 ) be a random hexagonal chain graph with n hexagons. If p 1 > 0 , then for each n 1 ,

E [ M ( H n ( p 1 , p 2 ) ) ] = 2 s 3 s t t n 1 2 t 3 s t s n 1 ,

where t = 2 p 1 + p 1 2 + 4 p 1 2 and s = 2 p 1 p 1 2 + 4 p 1 2 .

If p 1 = 0 , then, for each n 1 ,

E [ M ( H n ( 0 , 1 ) ) ] = 1 + n .

Corollary 3.4

Let P C n r ( p 1 , p 2 , , p k 1 ) be a random even-polygonal chain with n polygons. If k is even and i = 1 k 2 2 p 2 i < 1 (or k is odd and i = 1 k 1 2 p 2 i < 1 ), then

lim n E [ M ( P C n r ( p 1 , p 2 , , p k 1 ) ) ] E [ M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) ] = s ,

where

s = 1 + i = 1 k 2 2 p 2 i 3 i = 1 k 2 2 p 2 i 2 4 2 , i f k i s e v e n ; 1 + i = 1 k 1 2 p 2 i ( 3 i = 1 k 1 2 p 2 i ) 2 4 2 , i f k i s o d d .

Proof

Keeping the notation defined in Theorem 3.2, we obtain that

E [ M ( P C n r ( p 1 , p 2 , , p k 1 ) ) ] E [ M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) ] = k 1 s n 1 + k 2 t n 1 k 1 s n 2 + k 2 t n 2 = s + t k 2 k 1 t s n 2 1 + k 2 k 1 t s n 2 .

Since s > t , it follows that

lim n E [ M ( P C n r ( p 1 , p 2 , , p k 1 ) ) ] E [ M ( P C n 1 r ( p 1 , p 2 , , p k 1 ) ) ] = s .

Let C n r be the set of all even-polygonal chains with n polygons. The average value of the number for the perfect matchings of C n r is defined by:

M a v r ( C n r ) = 1 C n r P C n r C n r M ( P C n r ) .

Actually, this is the population mean of the number of perfect matchings of all elements in C n r . Since every element occurring in C n r has the same probability, we have p 1 = p 2 = = p k 1 = 1 k 1 . Thus, we may apply Theorem 3.2 by putting p 1 = p 2 = = p k 1 = 1 k 1 and obtain the following result.

Theorem 3.5

The average value of the number for the perfect matchings with respect to C n r is

M a v r ( C n r ) = 2 3 2 n 1 .

Proof

By Theorem 3.2, we obtain s = 3 2 and t = 0 . Hence, we have

M avr ( C n r ) = 0 3 3 2 0 3 2 n 1 = 2 3 2 n 1 .

Acknowledgements

We are grateful to the reviewers for providing us the valuable suggestions. This work was partially completed while the first author visited School of Mathematics and Statistics, Lanzhou University.

  1. Funding information: This work was supported by Natural Science Foundation of China (Nos. 12061007, 12071194, 11961067, and 11571155) and Natural Science Found of Fujian Province (No. 2020J01844).

  2. Author contributions: S. L. Wei and Y. D. Feng contributed to the supervision, methodology, validation, project administration, and formal analysis. X. L. Ke and J. W. Huang contributed to the investigation, resources, and some computations and wrote the initial draft of this article, which was investigated and approved by Y. D. Feng and S. L. Wei who wrote the final draft. All authors have read and agreed to the published version of the manuscript.

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

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

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Received: 2022-11-11
Revised: 2023-09-18
Accepted: 2023-10-19
Published Online: 2023-12-06

© 2023 the author(s), published by De Gruyter

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

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