All graphs considered are finite, undirected, loopless and without multiple edges. The terminology and nomenclature of [1] will be used. Throughout this paper, G will denote a graph with vertex set V ( G ) = { u , v , v 1 , v 2 , ⋯ , v n − 2 } and edge set E ( G ) . By G − u denote the induced subgraph obtained from G by deleting vertex u together with its incident edges and by G − e the edge-induced subgraph obtained from G by deleting edge e. As usual, use P n and C n to denote the path and cycle on n vertices, respectively.

Let m ( G , k ) be the number of k-matchings in graph G. The matching polynomial of a graph G is defined in [2] as

α ( G , λ ) = ∑ k ≥ 0 ( − 1 ) k m ( G , k ) λ n − 2 k . (1)

where m ( G , 0 ) = 1 and m ( G , k ) ≥ 0 for all k = 1 , 2 , ⋯ , ⌊ n 2 ⌋ .

Gutman and Wager defined the quasi-order “ ≽ ” of two graphs G and H as follows:

If G and H have the matching polynomials in the form (1), then the quasi- order “ ≽ ” is defined by

G ≽ H ⇔ m ( G , k ) ≥ m ( H , k )     for   all   k = 0,1, ⋯ , ⌊ n / 2 ⌋ . (2)

In particular, if G ≽ H and m ( G , k ) > m ( H , k ) for some k, then we write G ≻ H .

We call G and H matching-equivalent if both G ≽ H and H ≽ G hold and denoted by G ~ H . Further, Gutman and Wagner introduced the concept of matching energy M E ( G ) of a graph G in [3] and defined in different expressions as follows:

M E ( G ) = 2 π ∫ 0 ∞ 1 x 2 l n [ ∑ k ≥ 0     m ( G , k ) x 2 k ] d x . (3)

and

M E ( G ) = ∑ i = 1 n | λ i |

where λ 1 , λ 2 , ⋯ , λ n be the roots of matching polynomial of graph G.

The matching energy M E ( G ) of a graph G is an important index, which is widely used in the field of molecular orbital theory. There are many literatures about this parameter. See [4] - [14] .

By the above definitions, it is immediately to get

G ≽ H ⇒ M E ( G ) ≥ M E ( H )   and   G ≻ H ⇒ M E ( G ) > M E ( H ) . (4)

In fact, this property provide an important technique to determine the order relation of the matching energy for graphs. In this paper, we discuss the order of the matching energy for circum graph with chord.

Let C n = u v v 1 v 2 ⋯ v n − 2 u be a cycle with order n, the circum graph with chord is obtained by adding one edge u v i for some i ∈ { 1,2, ⋯ , n − 3 } to C n , which is denoted by G ( n ; i , n − 2 − i ) and simplified as G u ~ v i .

Lemma 1. [2] Let e = u v be an edge of graph G and k ≥ 1 . Then m ( G , k ) = Math_39#.

By Lemma 1, it is easy to get

Lemma 2. Let e be an edge of graph G. Then M E ( G − e ) < M E ( G ) .

Lemma 3. [15] Let u be a vertex of graph G. Then m ( G , k ) = m ( G − u , k ) + ∑ v     m ( G − u − v , k − 1 ) , where the summation goes over all vertices v adjacent to the vertex u.

Lemma 4. [16] Let n = 4 k + i , i ∈ { 0 , 1 , 2 , 3 } for k ≥ 1 . Then P n ≻ P 2 ∪ P n − 2 ≻ P 4 ∪ P n − 4 ≻ ⋯ ≻ P 2 k ∪ P n − 2 k ≻ P 2 k + 1 ∪ P n − 2 k − 1 ≻ P 2 k − 1 ∪ P n − 2 k + 1 ≻ ⋯ ≻ P 3 ∪ P n − 3 ≻ P 1 ∪ P n − 1 .

By the definition of graph G u ~ v i , we can immediately get

Lemma 5. G u ~ v i ~ G u ~ v n − 2 − i for i ∈ { 1,2, ⋯ , n − 3 } .

Proof. Since G u ~ v i = G ( n ; i , n − 2 − i ) and G u ~ v n − 2 − i = G ( n ; , n − 2 − i , i ) , so we get G u ~ v i ~ G u ~ v n − 2 − i .,

Theorem 6. Let G u ~ v i be a circum graph with chord and n is an even. Then G u ~ v 1 ≺ G u ~ v 3 ≺ ⋯ ≺ G u ~ v n − 2 2 − 1 ≺ G u ~ v n − 2 2 ≺ G u ~ v n − 2 2 − 2 ≺ ⋯ ≺ G u ~ v 4 ≺ G u ~ v 2 .

Proof. First we consider the graph G u ~ v s and G u ~ v s − 2 . Since n is even, by Lemma 5, we only consider 1 ≤ s ≤ n − 2 2 . By Lemma 3, we obtain that

m ( G u ~ v s , k ) = m [ G ( n ; s , n − 2 − s ) , k ] = m ( P s + ( n − 2 − s ) + 1 , k ) + m ( P ( s − 1 ) + ( n − 2 − s ) + 1 , k − 1 )     + m ( P s + ( n − 2 − s − 1 ) + 1 , k − 1 ) + m ( P s ∪ P n − 2 − s , k − 1 ) = m ( P n − 1 , k ) + 2 m ( P n − 2 , k − 1 ) + m ( P s ∪ P n − 2 − s , k − 1 ) (5)

Similarly,

m ( G u ~ v s − 2 , k ) = m [ G ( n ; s − 2 , n − s ) , k ] = m ( P ( s − 2 ) + ( n − s ) + 1 , k ) + m ( P ( s − 2 − 1 ) + ( n − s ) + 1 , k − 1 )     + m ( P ( s − 2 ) + ( n − s − 1 ) + 1 , k − 1 ) + m ( P s − 2 ∪ P n − s , k − 1 ) = m ( P n − 1 , k ) + 2 m ( P n − 2 , k − 1 ) + m ( P s − 2 ∪ P n − s , k − 1 ) (6)

Based on (5) and (6), we immediately get m ( G u ~ v s , k ) − m ( G u ~ v s − 2 , k ) = m ( P s ∪ P n − 2 − s , k − 1 ) − m ( P s − 2 ∪ P n − s , k − 1 ) .

Case 1. s is even.

By Lemma 4, we can obtain that P s ∪ P n − 2 − s ≺ P s − 2 ∪ P n − s . Thus, for some k, there be m ( G u ~ v s − 2 , k ) > m ( G u ~ v s , k ) . This means that G u ~ v 2 ≻ G u ~ v 4 ≻ G u ~ v 6 ≻ ⋯

≻ G u ~ v n − 2 2 .

Case 2. s is odd.

By Lemma 4, using a similar argument as in the previous proof we conclude that P s ∪ P n − 2 − s ≻ P s − 2 ∪ P n − s . Thus, for some k, there be m ( G u ~ v s , k ) > m ( G u ~ v s − 2 , k ) . This imply that G u ~ v n − 2 2 − 1 ≻ G u ~ v n − 2 2 − 3 ≻ ⋯ ≻ G u ~ v 3 ≻ G u ~ v 1 .

For graph G u ~ v n − 2 2 and G u ~ v n − 2 2 − 1 , we get that

m ( G u ~ v n − 2 2 , k ) − m ( G u ~ v n − 2 2 − 1 , k ) = m ( P n − 2 2 ∪ P n − 2 2 , k − 1 ) − m ( P n − 2 2 − 1 ∪ P n − 2 2 + 1 , k − 1 )

Repeating the same argument as in the previous proof, combine the fact n is even, we have P n − 2 2 ∪ P n − 2 2 ≻ P n − 2 2 − 1 ∪ P n − 2 2 + 1 . Thus G u ~ v n − 2 2 ≻ G u ~ v n − 2 2 − 1 . Sum up all, we get G u ~ v 1 ≺ G u ~ v 3 ≺ ⋯ ≺ G u ~ v n − 2 2 − 1 ≺ G u ~ v n − 2 2 ≺ G u ~ v n − 2 2 − 2 ≺ G u ~ v 2 . ,

Theorem 7. Let n is an odd number. Then

1) If ⌊ n − 2 2 ⌋ is also odd, then G u ~ v 2 ≻ G u ~ v 4 ≻ G u ~ v ⌊ n − 2 2 ⌋ − 1 ≻ G u ~ v ⌊ n − 2 2 ⌋ ≻ G u ~ v ⌊ n − 2 2 ⌋ − 2 ≻ ⋯ ≻ G u ~ v 3 ≻ G u ~ v 1 ;

2) If ⌊ n − 2 2 ⌋ is even, then G u ~ v 2 ≻ G u ~ v 4 ≻ G u ~ v ⌊ n − 2 2 ⌋ ≻ G u ~ v ⌊ n − 2 2 ⌋ − 1 ≻ G u ~ v ⌊ n − 2 2 ⌋ − 3 ≻ ⋯ ≻ G u ~ v 3 ≻ G u ~ v 1 .

Proof. First consider the graph G u ~ v s and G u ~ v s − 2 . Since n is odd, similar as Lemma 5, we only consider 1 ≤ s ≤ ⌊ n − 2 2 ⌋ . By Lemma 1, we get m ( G u ~ v s , k ) − m ( G u ~ v s − 2 , k ) = m ( P s ∪ P n − 2 − s , k − 1 ) − m ( P s − 2 ∪ P n − s , k − 1 ) .

Case 1. s is even.

By Lemma 4, we have P s ∪ P n − 2 − s ≺ P s − 2 ∪ P n − s . Thus, m ( G u ~ v s − 2 , k ) > Math_87#.

If ⌊ n − 2 2 ⌋ is odd, then G u ~ v 2 ≻ G u ~ v 4 ≻ G u ~ v 6 ≻ ⋯ ≻ G u ~ v ⌊ n − 2 2 ⌋ − 1 .

If ⌊ n − 2 2 ⌋ is even, then G u ~ v 2 ≻ G u ~ v 4 ≻ G u ~ v 6 ≻ ⋯ ≻ G u ~ v n − 2 2 .

Case 2. s is odd.

By Lemma 4, we have P s ∪ P n − 2 − s ≻ P s − 2 ∪ P n − s . Thus, m ( G u ~ v s , k ) > m ( G u ~ v s − 2 , k ) .

If ⌊ n − 2 2 ⌋ is odd, then G u ~ v ⌊ n − 2 2 ⌋ ≻ G u ~ v ⌊ n − 2 2 ⌋ − 2 ≻ ⋯ ≻ G u ~ v 3 ≻ G u ~ v 1 .

If ⌊ n − 2 2 ⌋ is even, then G u ~ v ⌊ n − 2 2 ⌋ − 1 ≻ G u ~ v ⌊ n − 2 2 ⌋ − 3 ≻ ⋯ ≻ G u ~ v 3 ≻ G u ~ v 1 .

Based on the above analysis, if ⌊ n − 2 2 ⌋ is odd,

m ( G u ~ v ⌊ n − 2 2 ⌋ − 1 , k ) − m ( G u ~ v ⌊ n − 2 2 ⌋ , k ) = m [ G ( n ; ⌊ n − 2 2 ⌋ − 1, ⌈ n − 2 2 ⌉ + 1 ) , k ] − m [ G ( n ; ⌊ n − 2 2 ⌋ , ⌈ n − 2 2 ⌉ + 1 ) , k ] = m ( P ⌊ n − 2 2 ⌋ − 1 ∪ P ⌈ n − 2 2 ⌉ + 1 , k − 1 ) − m ( P ⌊ n − 2 2 ⌋ ∪ P ⌈ n − 2 2 ⌉ , k − 1 )

By lemma 4, P ⌊ n − 2 2 ⌋ − 1 ∪ P ⌈ n − 2 2 ⌉ + 1 ≻ P ⌊ n − 2 2 ⌋ ∪ P ⌈ n − 2 2 ⌉ . Thus for some k, we have m ( G u ~ v ⌊ n − 2 2 ⌋ − 1 , k ) > m ( G u ~ v ⌊ n − 2 2 ⌋ , k ) . This means G u ~ v ⌊ n − 2 2 ⌋ − 1 ≻ G u ~ v ⌊ n − 2 2 ⌋ .

If ⌊ n − 2 2 ⌋ is even, then

m ( G u ~ v ⌊ n − 2 2 ⌋ , k ) − m ( G u ~ v ⌊ n − 2 2 ⌋ − 1 , k ) = m [ G ( n ; ⌊ n − 2 2 ⌋ , ⌈ n − 2 2 ⌉ ) , k ] − m [ G ( n ; ⌊ n − 2 2 ⌋ − 1 , ⌈ n − 2 2 ⌉ + 1 ) , k ] = m ( P ⌊ n − 2 2 ⌋ ∪ P ⌈ n − 2 2 ⌉ , k − 1 ) − m ( P ⌊ n − 2 2 ⌋ − 1 ∪ P ⌈ n − 2 2 ⌉ + 1 , k − 1 )

By Lemma 4, P ⌊ n − 2 2 ⌋ ∪ P ⌈ n − 2 2 ⌉ ≻ P ⌊ n − 2 2 ⌋ − 1 ∪ P ⌈ n − 2 2 ⌉ + 1 . Thus for some k, we have m ( G u ~ v ⌊ n − 2 2 ⌋ , k ) > m ( G u ~ v ⌊ n − 2 2 ⌋ − 1 , k ) . This means that G u ~ v ⌊ n − 2 2 ⌋ ≻ G u ~ v ⌊ n − 2 2 ⌋ − 1 . Sum up all, for n is an odd, if ⌊ n − 2 2 ⌋ is also odd, then G u ~ v 2 ≻ G u ~ v 4 ≻ G u ~ v ⌊ n − 2 2 ⌋ − 1 Math_110#;

If ⌊ n − 2 2 ⌋ is an even, then G u ~ v 2 ≻ G u ~ v 4 ≻ G u ~ v ⌊ n − 2 2 ⌋ ≻ G u ~ v ⌊ n − 2 2 ⌋ − 1 ≻ G u ~ v ⌊ n − 2 2 ⌋ − 3 ≻ ⋯ ≻ G u ~ v 3 ≻ G u ~ v 1 .

By Theorems 6 and 7, we immediately get our main result as follow.

Theorem 8. Let G u ~ v i ( i = 1 , 2 , ⋯ , n − 3 ) be a circum graphs with chord.

1) If n is an even, then

M E ( G u ~ v 1 ) < M E ( G u ~ v 3 ) < ⋯ < M E ( G u ~ v ⌊ n − 2 2 ⌋ ) < M E ( G u ~ v ⌊ n − 2 2 ⌋ − 1 ) < M E ( G u ~ v ⌊ n − 2 2 ⌋ − 3 ) < ⋯ < M E ( G u ~ v 2 ) < M E ( G u ~ v 2 ) (7)

2) If n and ⌊ n − 2 2 ⌋ are both odd, then

M E ( G u ~ v 1 ) < M E ( G u ~ v 3 ) < ⋯ < M E ( G u ~ v ⌊ n − 2 2 ⌋ − 2 ) < M E ( G u ~ v ⌊ n − 2 2 ⌋ ) < M E ( G u ~ v ⌊ n − 2 2 ⌋ − 1 ) < ⋯ < M E ( G u ~ v 4 ) < M E ( G u ~ v 2 ) (8)

3) If n is an odd and ⌊ n − 2 2 ⌋ is an even, then

M E ( G u ~ v 1 ) < M E ( G u ~ v 3 ) < ⋯ < M E ( G u ~ v ⌊ n − 2 2 ⌋ − 3 ) < M E ( G u ~ v ⌊ n − 2 2 ⌋ − 1 ) < M E ( G u ~ v ⌊ n − 2 2 ⌋ ) < ⋯ < M E ( G u ~ v 4 ) < M E ( G u ~ v 2 ) (9)

In this paper, we determine the quasi-order relation on the matching energy for circum graph with one chord. If the chord here can be see P 2 . Then the general case, determining the quasi-order relation on the matching energy for circum graph with one generalized chord P k for 2 ≤ k ≤ n − 3 is more meaningful.

Sincere thanks to the members of JAMP for their professional performance, and special thanks to managing editor for a rare attitude of high quality. This research supported by NSFC (11561056, 11661066) and QHAFP (2017-ZJ-701).

Zhao, N. and Li, Y.K. (2017) The Quasi-Order of Matching Energy of Circum Graph with Chord. Applied Mathematics, 8, 1180-1185. https://doi.org/10.4236/am.2017.88088