This paper only considers finite undirected simple graphs. Let G be a graph with n vertices. The matching of G means a spanning subgraph of G, and each of its connected branches is either an isolated vertex or an isolated edge. t-matching means matching that there are t edges. Matching polynomial of graph G is defined as follows in [2]:

μ ( G , x ) = ∑ t ≥ 0 ( − 1 ) t α t ( G ) x n − 2 t , (1)

where α t ( G ) is number of t-matching for G.

If graphG and graphH satisfy μ ( G , x ) = μ ( H , x ) , then we say G and H are matching equivalence, denoted as G ~ H .

Set δ ( G ) which denotes the number of matching equivalent graphs of all different isomorphisms for graphG, if δ ( G ) = 1 , we say graphG is a unique matching. In [2], the authors gave some elegant properties of matching polynomials, and proved that to find the matching polynomial of a graph is an NP-problem. Thus, in [1], the authors studied the number of matching equivalent graphs of some vertices and some cycle-union graphs. It turns out that the problem is not simple. In this paper, we study the number of matching equivalent for the union graph of vertices and cycles, that is δ ( s K 1 ∪ t 1 C 9 ∪ t 2 C 15 ) .

Throughout the paper, K₁ denotes an isolated vertex, P n ( n ≥ 2 ) denotes the path including n vertices; C m ( m ≥ 3 ) denotes a cycle that including m vertices; T i , j , k denotes a tree that has only a 3-degree vertex, three 1-degree vertex, and the distance between this 3-degree vertex and three 1-degree vertex is i , j , k separately; D n ( n ≥ 4 ) denotes a graph that produced by bonding a vertex on a triangle to an end of a path P n − 2 ; nG denotes disjoint union of n graph G.

Lemma 2.1 [3] Suppose graph G has k connected component: G 1 , G 2 , ⋯ , G k , then μ ( G , x ) = ∏ i = 1 k μ ( G i , x ) , the roots of matching polynomials are all real numbers, denote M ( G ) as the maximum root of μ ( G , x ) .

Lemma 2.2 [3] Suppose G is connected graph, then M ( G ) < 2 if and only if G ∈ Γ = { K 1 , P n , T 1 , 1 , n , T 1 , 2 , 2 , T 1 , 2 , 3 , T 1 , 2 , 4 , C n , D 4 } .

Lemma 2.3 [3] (i) M ( C m ) = M ( T 1 , 1 , m − 2 ) = M ( P 2 m − 1 ) .

(ii) M ( C 6 ) = M ( T 1 , 1 , 4 ) = M ( T 1 , 2 , 2 ) = M ( D 4 ) = M ( P 11 ) .

(iii) M ( C 9 ) = M ( T 1 , 1 , 7 ) = M ( T 1 , 2 , 3 ) = M ( P 17 ) .

(iv) M ( C 15 ) = M ( T 1 , 1 , 13 ) = M ( T 1 , 2 , 4 ) = M ( P 29 ) .

Lemma 2.4 [4] [5] (i) The matching equivalent graph of K 1 ∪ C m ( m ≠ 6 , 9 , 15 ) is K 1 ∪ C m , T 1 , 1 , m − 2 .

(ii) The matching equivalent graph of K 1 ∪ C 6 is: K 1 ∪ C 6 , T 1 , 1 , 4 , P 3 ∪ D 4 .

(iii) The matching equivalent graph of K 1 ∪ C 9 is K 1 ∪ C 9 , T 1 , 1 , 7 , C 3 ∪ T 1,2,3 .

(iv) The matching equivalent graph of K 1 ∪ C 15 is: K 1 ∪ C 15 ~ C 3 ∪ C 5 ∪ T 1,2,4 .

Lemma 2.5 [6] (i) P 2 m + 1 ~ P m ∪ C m + 1 , ( m ≥ 2 ) .

(ii) T 1 , 1 n ~ K 1 ∪ C n + 2 .

(iii) T 1 , 2 , 2 ~ P 2 ∪ D 4 .

(iv) K 1 ∪ C 6 ~ P 3 ∪ D 4 .

(v) K 1 ∪ C 9 ~ C 3 ∪ T 1,2,3 .

(vi) K 1 ∪ C 15 ~ C 3 ∪ C 5 ∪ T 1,2,4 .

Lemma 2.6 [1] Suppose G = s K 1 or t 1 C m 1 ∪ ⋯ ∪ t k C m k , then G is a unique matching, that is δ ( G ) = 1 .

Lemma 2.7 [1] Suppose

G = s K 1 ∪ t 1 C m 1 ∪ t 2 C m 2 ∪ ⋯ ∪ t k C m k , m i ≠ 6   ( i = 1 , 2 , ⋯ , k ) ,

then all matching equivalent graphs of G do not contain road branches.

Lemma 2.8 [1]

(i) G = s K 1 ∪ a P 3 ∪ t C 6 , then all matching equivalent graphs ofG do not contain P 11 branch and also T 1 , 2 , 2 [7],

(ii) δ ( s K 1 ∪ a P 3 ∪ t C 6 ) = δ ( s K 1 ∪ t C 6 ) .

Lemma 2.9 [1] [7] If m ≠ 6 , 9 , 15 , then

δ ( s K 1 ∪ t C 6 ) = min { s , t } + 1 . (2)

Lemma 2.10 [1] [8] If m i ≠ 6 , 9 , 15 ( i = 1 , 2 ) , then

δ ( s K 1 ∪ t 1 C m 1 ∪ t 2 C m 2 ) = ∑ i = 0 r min { s − i , t 1 } + r + 1 , where

r = min { s , t 2 } . (3)

Lemma 3.1 δ ( s K 1 ∪ t 1 C 3 ∪ t 2 C 5 ∪ t 3 C 9 ) = ∑ j = 0 r ∑ i = j r δ ( s − i , t 1 + ( i − j ) , t 2 ) ,

where

r = min { s , t 3 } . (4)

Proof. For simplicity, denote δ ( s K 1 ∪ t 1 C 3 ∪ t 2 C 5 ∪ t 3 C 9 ) = δ ( s , t 1 , t 2 , t 3 ) .

Set H ~ s K 1 ∪ t 1 C 3 ∪ t 2 C 5 ∪ t 3 C 6 , by lemma 2.3 (iii) and lemma 2.7, we know H contains connected component C 9 , T 1 , 1 , 7 or T 1 , 2 , 3 .

1) IfH contains C 9 , by H = C 9 ∪ H 2 ~ s K 1 ∪ t 1 C 3 ∪ t 2 C 5 ∪ t 3 C 9 , we know H 2 ~ s K 1 ∪ t 1 C 3 ∪ t 2 C 5 ∪ ( t 3 − 1 ) C 9 .

Such H 2 has a total of δ ( s , t 1 , t 2 , t 3 − 1 ) .

2) IfH contains T 1 , 1 , 7 , by H = T 1 , 1 , 7 ∪ H 2 ~ s K 1 ∪ t 1 C 3 ∪ t 2 C 5 ∪ t 3 C 9 and lemma 2.5 (ii), we know

H 2 ~ ( s − 1 ) K 1 ∪ t 1 C 3 ∪ t 2 C 5 ∪ ( t 3 − 1 ) C 9 ,

Such H 2 has a total of δ ( s − 1 , t 1 , t 2 , t 3 − 1 ) .

3) IfH contains T 1 , 2 , 3 , by H = T 1 , 2 , 3 ∪ H 2 ~ s K 1 ∪ t 1 C 3 ∪ t 2 C 5 ∪ t 3 C 9 and lemma 2.5(v), we get

H 2 ~ ( s − 1 ) K 1 ∪ ( t 1 + 1 ) C 3 ∪ t 2 C 5 ∪ ( t 3 − 1 ) C 9 ,

Such H 2 has a total of δ ( s − 1 , t 1 + 1 , t 2 , t 3 − 1 ) .

4) IfH contains C 9 and T 1 , 1 , 7 simultaneously, such H 2 has a total of δ ( s − 1 , t 1 , t 2 , t 3 − 2 ) .

5) IfH contains C 9 and T 1 , 2 , 3 simultaneously, such H 2 has a total of δ ( s − 1 , t 1 + 1 , t 2 , t 3 − 2 ) .

6) IfH contains T 1 , 1 , 7 and T 1 , 2 , 3 simultaneously, such H 2 has a total of δ ( s − 2 , t 1 + 1 , t 2 , t 3 − 2 ) .

7) IfH contains C 9 , T 1 , 1 , 7 and T 1 , 2 , 3 simultaneously, such H 2 has a total of δ ( s − 2 , t 1 + 1 , t 2 , t 3 − 3 ) .

Thus,

δ ( s , t 1 , t 2 , t 3 ) = δ ( s , t 1 , t 2 , t 3 − 1 ) + δ ( s − 1 , t 1 , t 2 , t 3 − 1 ) + δ ( s − 1 , t 1 + 1 , t 2 , t 3 − 1 )   − δ ( s − 1 , t 1 , t 2 , t 3 − 2 ) − δ ( s − 1 , t 1 + 1 , t 2 , t 3 − 2 )   − δ ( s − 2 , t 1 + 1 , t 2 , t 3 − 2 ) + δ ( s − 2 , t 1 + 1 , t 2 , t 3 − 3 )

Then,

δ ( s , t 1 , t 2 , t 3 ) − δ ( s − 1 , t 1 , t 2 , t 3 − 1 ) − δ ( s − 1 , t 1 + 1 , t 2 , t 3 − 1 )   + δ ( s − 2 , t 1 + 1 , t 2 , t 3 − 2 ) = δ ( s , t 1 , t 2 , t 3 − 1 ) − δ ( s − 1 , t 1 , t 2 , t 3 − 2 ) − δ ( s − 1 , t 1 + 1 , t 2 , t 3 − 2 )   + δ ( s − 2 , t 1 + 1 , t 2 , t 3 − 3 )

Repeat the application of the above formula, we obtain

δ ( s , t 1 , t 2 , t 3 ) − δ ( s − 1 , t 1 , t 2 , t 3 − 1 ) − δ ( s − 1 , t 1 + 1 , t 2 , t 3 − 1 )   + δ ( s − 2 , t 1 + 1 , t 2 , t 3 − 2 ) = δ ( s , t 1 , t 2 , 2 ) − δ ( s − 1 , t 1 , t 2 , 1 ) − δ ( s − 1 , t 1 + 1 , t 2 , 1 ) + δ ( s − 2 , t 1 + 1 , t 2 , 0 ) = δ ( s , t 1 , t 2 , 1 ) − δ ( s − 1 , t 1 , t 2 , 0 ) − δ ( s − 1 , t 1 + 1 , t 2 , 0 ) = δ ( s , t 1 , t 2 , 0 ) = δ ( s , t 1 , t 2 )

Thus,

δ ( s , t 1 , t 2 , t 3 ) − δ ( s − 1 , t 1 , t 2 , t 3 − 1 ) = δ ( s − 1 , t 1 + 1 , t 2 , t 3 − 1 ) − δ ( s − 2 , t 1 + 1 , t 2 , t 3 − 2 ) + δ ( s , t 1 , t 2 ) = δ ( s − 2 , t 1 + 2 , t 2 , t 3 − 2 ) − δ ( s − 3 , t 1 + 2 , t 2 , t 3 − 3 )       + δ ( s − 1 , t 1 + 1 , t 2 ) + δ ( s − 1 , t 1 + 1 , t 2 ) = ⋯ = ∑ i = 0 r δ ( s − i , t 1 + i , t 2 )

δ ( s , t 1 , t 2 , t 3 ) − δ ( s − 1 , t 1 , t 2 , t 3 − 1 ) = ∑ i = 0 r δ ( s − i , t 1 + i , t 2 ) (1')

δ ( s − 1 , t 1 , t 2 , t 3 − 1 ) − δ ( s − 2 , t 1 , t 2 , t 3 − 2 ) = ∑ i = 1 r δ ( s − i , t 1 + ( i − 1 ) , t 2 ) (2')

⋯

δ ( s − ( r − 1 ) , t 1 , t 2 , t 3 − ( r − 1 ) ) − δ ( s − r , t 1 , t 2 , t 3 − r ) = ∑ i = r − 1 r δ ( s − i , t 1 + i − ( r − 1 ) , t 2 ) (r')

δ ( s − r , t 1 , t 2 , t 3 − r ) = ∑ i = r r δ ( s − i , t 1 + ( i − r ) , t 2 ) (r+1')

Add (1'), (2'), ⋯ , (r+1') together, we get

δ ( s , t 1 , t 2 , t 3 ) = ∑ j = 0 r ∑ i = j r δ ( s − i , t 1 + ( i − j ) , t 2 ) . +

Theorem 3.1

δ ( s K 1 ∪ t 1 C 9 ∪ t 2 C 15 ) = ∑ j = 0 r ∑ i = j r δ ( s − i , i − j , i − j , t 1 ) . (5)

Proof. For simplicity, denote

δ ( s K 1 ∪ t C 3 ∪ t ′ C 5 ∪ t 1 C 9 ∪ t 2 C 15 ) = δ ( s , t , t ′ t 1 , t 2 ) .

Suppose H ~ G , by lemma 2.3(iv) and lemma 2.7, we know H contains connected component C 15 , T 1 , 1 , 13 or T 1 , 2 , 4 .

1) IfH contains C 15 , by H = C 15 ∪ H 2 ~ s K 1 ∪ t 1 C 9 ∪ t 2 C 15 we know H 2 ~ s K 1 ∪ t 1 C 9 ∪ ( t 2 − 1 ) C 15 . Such H 2 has a total of δ ( s , 0 , 0 , t 1 , t 2 − 1 ) .

2) IfH contains T 1 , 1 , 13 , by H = T 1 , 1 , 13 ∪ H 2 ~ G = s K 1 ∪ t 1 C 9 ∪ t 2 C 15 and lemma 2.5(ii), we get H 2 ~ ( s − 1 ) K 1 ∪ t 1 C 9 ∪ ( t 2 − 1 ) C 15 . Such H 2 has a total of δ ( s − 1 , 0 , 0 , t 1 , t 2 − 1 ) .

3) IfH contains T 1 , 2 , 4 , by H = T 1 , 2 , 4 ∪ H 2 ~ G = s K 1 ∪ t 1 C 9 ∪ t 2 C 15 and lemma 2.5(vi), we get H 2 ~ ( s − 1 ) K 1 ∪ C 3 ∪ C 5 ∪ t 1 C 9 ∪ ( t 2 − 1 ) C 15 , such H 2 has a total of δ ( s − 1 , 1 , 1 , t 1 , t 2 − 1 ) .

4) If H contains C 15 and T 1 , 1 , 13 , such H 2 has a total of δ ( s − 1 , 0 , 0 , t 1 , t 2 − 2 ) .

5) IfH contains C 15 and T 1 , 2 , 4 , such H 2 has a total of δ ( s − 1 , 1 , 1 , t 1 , t 2 − 2 ) .

6) If H contains T 1 , 1 , 13 and T 1 , 2 , 4 , such H 2 has a total of δ ( s − 2 , 1 , 1 , t 1 , t 2 − 2 ) .

7) If H contains C 15 , T 1 , 1 , 13 and T 1 , 2 , 4 , such H 2 has a total of δ ( s − 2 , 1 , 1 , t 1 , t 2 − 3 ) .

Then

δ ( s , 0 , 0 , t 1 , t 2 ) = δ ( s , 0 , 0 , t 1 , t 2 − 1 ) + δ ( s − 1 , 0 , 0 , t 1 , t 2 − 1 )   + δ ( s − 1 , 1 , 1 , t 1 , t 2 − 1 ) − δ ( s − 1 , 0 , 0 , t 1 , t 2 − 2 )   − δ ( s − 1 , 1 , 1 , t 1 , t 2 − 2 ) − δ ( s − 2 , 1 , 1 , t 1 , t 2 − 2 )   + δ ( s − 2 , 1 , 1 , t 1 , t 2 − 3 )

Thus,

δ ( s , 0 , 0 , t 1 , t 2 ) − δ ( s − 1 , 0 , 0 , t 1 , t 2 − 1 ) − δ ( s − 1 , 1 , 1 , t 1 , t 2 − 1 )   + δ ( s − 2 , 1 , 1 , t 1 , t 2 − 2 ) = δ ( s , 0 , 0 , t 1 , t 2 − 1 ) − δ ( s − 1 , 0 , 0 , t 1 , t 2 − 2 ) − δ ( s − 1 , 1 , 1 , t 1 , t 2 − 2 )   + δ ( s − 2 , 1 , 1 , t 1 , t 2 − 3 )

Repeat the application of the above formula, we obtain

δ ( s , 0 , 0 , t 1 , t 2 ) − δ ( s − 1 , 0 , 0 , t 1 , t 2 − 1 ) − δ ( s − 1 , 1 , 1 , t 1 , t 2 − 1 )   + δ ( s − 2 , 1 , 1 , t 1 , t 2 − 2 ) = δ ( s , 0 , 0 , t 1 , 2 ) − δ ( s − 1 , 0 , 0 , t 1 , 1 ) − δ ( s − 1 , 1 , 1 , t 1 , 1 ) + δ ( s − 2 , 1 , 1 , t 1 , 0 ) = δ ( s , 0 , 0 , t 1 , 1 ) − δ ( s − 1 , 0 , 0 , t 1 , 0 ) − δ ( s − 1 , 1 , 1 , t 1 , 0 ) = δ ( s , 0 , 0 , t 1 )

So,

δ ( s , 0 , 0 , t 1 , t 2 ) − δ ( s − 1 , 0 , 0 , t 1 , t 2 − 1 ) = δ ( s − 1 , 1 , 1 , t 1 , t 2 − 1 ) − δ ( s − 2 , 1 , 1 , t 1 , t 2 − 2 ) + δ ( s , 0 , 0 , t 1 ) = δ ( s − 2 , 2 , 2 , t 1 , t 2 − 2 ) − δ ( s − 3 , 2 , 2 , t 1 , t 2 − 3 )     + δ ( s , 0 , 0 , t 1 ) + δ ( s − 1 , 1 , 1 , t 1 )

= δ ( s − 3 , 3 , 3 , t 1 , t 2 − 3 ) − δ ( s − 4 , 3 , 3 , t 1 , t 2 − 4 ) + δ ( s , 0 , 0 , t 1 )     + δ ( s − 1 , 1 , 1 , t 1 ) + δ ( s − 2 , 2 , 2 , t 1 ) = ⋯ = ∑ i = 0 r δ ( s − i , i , i , t 1 )

δ ( s , 0 , 0 , t 1 , t 2 ) − δ ( s − 1 , 0 , 0 , t 1 , t 2 − 1 ) = ∑ i = 0 r δ ( s − i , i , i , t 1 ) (1'')

δ ( s − 1 , 0 , 0 , t 1 , t 2 − 1 ) − δ ( s − 2 , 0 , 0 , t 1 , t 2 − 2 ) = ∑ i = 0 r δ ( s − i , i − 1 , i − 1 , t 1 ) (2'')

⋯

δ ( s − ( r − 1 ) , 0 , 0 , t 1 , t 2 − ( r − 1 ) ) − δ ( s − r , 0 , 0 , t 1 , t 2 − r ) = ∑ i = r − 1 r δ ( s − i , i − ( r − 1 ) , i − ( r − 1 ) , t 1 ) (r'')

δ ( s − r , 0 , 0 , t 1 , t 2 − r ) = ∑ i = r r δ ( s − i , i − r , i − r , t 1 ) (r+1'')

Add (1''), (2''), ⋯ , (r+1'') together, we get:

δ ( s , 0 , 0 , t 1 , t 2 ) = ∑ j = 0 r ∑ i = j r δ ( s − i , i − j , i − j , t 1 ) . +

Characterizing all graphs determined by a graph polynomial is an important subject in algebraic graph theory, among them, matching polynomial is considered to be a better algebraic tool. It is NP difficult to completely characterize the matched equivalent graphs of a class of graphs. In this paper, we study the number of matching equivalent graphs of some points and some cycli-union graphs, that is to say we calculate

δ ( s K 1 ∪ t 1 C 9 ∪ t 2 C 15 ) = ∑ j = 0 r ∑ i = j r δ ( s − i , i − j , i − j , t 1 ) .

This work is supported by the National Natural Science Foundation of China (No. 11561056), the Natural Science Foundation of Qinghai Province (2016-ZJ-914), Teaching Reform Research Project of Qinghai Nationalities University (2021-JYQN-005).

The author declares no conflicts of interest regarding the publication of this paper.

Wang, X.L. (2021) The Number of Matching Equivalent for the Union Graph of Vertices and Cycles. Applied Mathematics, 12, 471-476. https://doi.org/10.4236/am.2021.126032