Throughout this paper, we consider finite undirected simple graphs and follow the notation and terminology of [1].

Let G = ( V , E ) be a graph and we use δ ( G ) , Δ ( G ) and ω ( G ) to denote the minimum degree, the maximum degree and the number of components of the graph G , respectively. For a vertex v ∈ V ( G ) , we use N G ( v ) to denote the neighborhood of v in G . Let S be a subset of V ( G ) . We use G [ S ] to denote the subgraph of G induced on S , and also use P n , K n and C n to denote the path, the complete graph and the cycle of order n , respectively, where K 3 is called triangle. Let K m , n is the complete bipartite graph with partition sets of size m and n . We call K 1 , 3 and K 1 , k a claw and a star, respectively.

Let H be a given graph. A graph G is said to be H -free if G does not contain H as an induced subgraph. For a set of connected graphs ℋ . A graph G is said to be ℋ -free if G does not contain H as an induced subgraph for all H ∈ ℋ . The set ℋ was defined as the set of forbiddensubgraphs of G . We call ℋ a forbidden pair if | ℋ | = 2 . A subgraph containing all vertices of G is called a spanningsubgraph of G . If G contains a spanning cycle, then G is Hamiltonian. A subgraph of G is called even if its degrees are all positive even. A subgraph of G is called an even-factor if it is an even spanning subgraphs. A subgraph of G is called 2-factor if it is an even factor its degrees are all two. A graph is called supereulerian if it contains an even factor with exactly one component. For a subset X ⊆ E ( G ) , the contraction G / X is the graph obtained G by identifying the ends of each edge in X and then deleting the resulting loops. Note that the edges in E ( G / X ) can be regarded as edges in E ( G ) . If H is a subgraph, then we use G / H for G / E ( H ) . A graph H is a minor if it is isomorphic to the contraction image of a subgraph of G . A graph H is called an inducedminor if it is isomorphic to a contraction image of an induced subgraph of G .

In 1972, Richard M. Karp [2] proved that the Hamiltonian problem is NP-hard. After that, Pulleyblank pointed out in [3] that determining whether a graph is supereulerian is NP-hard, even when the problem is restricted to planar graphs. In most cases, when studying the Hamiltonian problem or the Supereulerian problem, we tend to start with the conditions of forbidden subgraphs.

In 1991, Bedrossian first gave all connected forbidden pairs for hamiltonicity of 2-connected graphs in [4]. In 1997, Faudree and Gould further proved this result of Bedrossian in [5] by completely characterizing all forbidden pairs for hamiltonicity 2-connected graphs with sufficiently large order.

Theorem 1.(Faudree and Gould [5]) Let R , S be connected graphs of order at least three other than P 3 . Then every 2-connected { R , S } -free graph with order at least n 0 is hamiltonian if and only if R = K 1 , 3 and S is an induced subgraph of P 6 , B 1 , 2 , N 1 , 1 , 1 , Z 3 .

Meanwhile, they also characterized the case of a forbidden subgraph for hamiltonicity of 2-connected graphs.

Theorem 2.(Faudree and Gould [5]) Let H be a connected graph of order at least 3 and G be a 2-connected graph. Then G is H -free implies G is hamiltonian if and only if H = P 3 .

A wheel W n is the graph obtained from the cycle C n = u 1 u 2 ⋯ u n , and an additional vertex u by joining u with u i for u ∈ { 1 , 2 , ⋯ , n } . Define W ′ n to be the graph obtained from W n by subdividing each edge of E ( C n ) once. Let W ″ 3 be the graph from C 7 and K 1 by joining K 1 to three non-adjacent vertices of C 7 . Let F 2 be the graph from C 4 and P 3 by identifying a vertex of C 4 with a endvertex of P 3 . Let Z i be the graph by identifying end vertices of one path of length i to the one vertex of triangle. Let B i , j be the graph by identifying end vertices of two disjoint paths of lengths i , j to the two vertices of triangle. Let N i , j , k be the graph by identifying end vertices of three disjoint paths of lengths i , j , k to the three vertices of the triangle. Here Z i , B i , j , N i , j , k with { i , j , k } ≥ 1 , see Figure 1.

In 1995, Lai [6] considered the same problem and obtained the following result.

Theorem 3.(Lai [6]) Let G be a 2-edge-connected graph. Then every 2-edge-connected graph induced subgraph of G is supereulerian if and only if G has no induced minor isomorphic to a member in W .

Figure 1. Some common induced subgraphs and W ′ 3 , W ′ 3 .

Over time, many researchers have shifted their focus from the study of Hamiltonian problems to the exploration of graph factorization problems. In [7], Tutte proved that a proved G has a 1-factor if and only if the deletion of an arbitrary vertex set S makes G − S have at most | S | odd branches. Later, Gallai in [8] began to facous on k -factors. In 2007, Plummer published a survey article on graph factorization in [9], which is widely regarded as the most authoritative literature on graph factorization problems.

In 2008, Xiong [10] gave a necessary and sufficient condition for a claw-free graph G to have an even factor.

Theorem 4.(Xiong [10]) Let G be a claw-free graph. Then G has an even factor if and only if δ ( G ) ≥ 2 and every odd branch-bond of G has an edge branch.

Furthermore, in [11], he extended the above result and characterized all forbidden subgraphs H such that every H -free graph G has an even factor if and only if δ ( G ) ≥ 2 and every odd branch-bond of G has an edge branch.

Theorem 5.(Xiong [11]) Let H be a connected graph of order at least three and G be a graph other than W 3 ' such that δ ( G ) ≥ 2 and every odd branch-bond of G has an edge branch. Then every H -free graph G has an even factor if and only if H is an induced subgraph of P 5 or T 2 , 1 , 1 .

It is easy to see that if there exists an even factor in a graph G , then G satisfies the following conditions:

1) δ ( G ) ≥ 2 ;

2) every odd branch-bond of G has an edge branch.

In 2017, Lv and Xiong characterized all forbidden pairs of connected subgraphs for supereulerian of 2-connected graphs with sufficiently large order in [12].

Theorem 6.(Lv and Xiong [12]) Let R , S be a pair of graphs and G be a 2-connected graph with n ≥ 10 . Then every { R , S } -free graph is supereulerian, if and only if (up to symmetry):

1) R = K 1 , 4 and S is an induced subgraph of P 5 ;

2) R = K 1 , 3 and S is an induced subgraph of P 7 , Z 4 , N 1 , 1 , 3 ;

3) R = C 4 and S is an induced subgraph of P 5 .

It is obviously that a supereulerian graph has an even factor and a Hamiltonian graph must be supereulerian. In other words, a necessary condition for a graph to be supereulerian is that it has an even factor and a necessary condition for a graph to be Hamiltonian is that it is supereulerian.

In order to illustrate some results better, we still need some definitions. A nontrivial path is called a branch if it has only internal vertices of degree two and end vertices of degree not two. The length of a branch is the number of its edges. Note that a branch of length one has no internal vertex, it is called an edgebranch. We denote by ℬ ( G ) the set of branches of G . For any subset S of ℬ ( G ) , we denote by G − S the subgraph obtained from G by deleting all edges and internal vertices of branches of S . A subset S of ℬ ( G ) is called a branchcut if G − S has more components than G . A minimal branch cut is called a branch-bond. A branch-bond is called odd if it has an odd numbers of branches.

In 2021, Yang, Du and Xiong [13] further considered the forbidden subgraphs for supereulerian of 2-edge-connected graphs satisfying that every odd branch bond of G has an edge branch, and characterized all forbidden subgraphs for supereulerian.

Theorem 7.(Yang, Du and Xiong [13]) Let H be a connected graph of order at least three. Then every 2-edge-connected H -free graph G satisfying that every odd branch bond of G has an edge branch is supereularian if and only if H is an induced subgraph of P 4 .

Theorem 8.(Yang, Du and Xiong [13]) Let R , S be a pair of graphs and G be a 2-edge-connected graph satisfying that every odd branch bond of G has an edge branch. Then every { R , S } -free graph G of order at least 11 is supereulerian if and only if { R , S } ≺ _ { T 1 , 1 , 2 , H 1 } , { T 1 , 1 , 2 , Z 6 } , { T 1 , 1 , 2 , B 1 , 5 } , { T 1 , 1 , 2 , B 3 , 3 } , { T 1 , 1 , 2 , N 1 , 2 , 3 } , { T 1 , 1 , 3 , Z 2 } or { P 5 , W } .

The number of components of an even factor of a graph is a key parameter characterizing the sparsity of the even factor: when the number of components is 1, the even factor degenerates into an even cycle; when the number of components is at least 3, the structure tends to be complex. An even factor with exactly two components serves as an important intermediate case connecting single even cycles and even factors with multiple components, and studying its existence conditions can fill the gap in the classification research on the number of components of even factors. Meanwhile, it can serve as an intermediate target for algorithm optimization, and can also be used to design more efficient algorithms for determining the existence of an arbitrary even factor, thereby having direct application value for network reliability.

Motivated by their results, we characterize all forbidden subgraphs for the existence of an even factor with exactly two components in a 2-edge-connected graph. The results as follows:

Theorem 9.Let H be a connected graph of order at least three, and G be a 2-edge-connected graph such that δ ( G ) ≥ 2 satisfying that every odd branch-bond of G has an edge branch. Then every H -free graph G has an even factor with exactly two components if and only if H is an induced subgraph of P 5 .

Theorem 10.Let R , S be connected graphs such that { R , S } ≇ { P 5 } . Let G be a graph of order at least 10 such that δ ( G ) ≥ 2 and every odd branch-bond of G has an edge branch. Then every { R , S } -free graph G has an even factor with exactly two components if and only if (up to symmetry):

1) R = F 2 and S is an induced subgraph of P 8 ;

2) R = T 2 , 1 , 1 and S is an induced subgraph of P 8 , Z 5 , B 2 , 3 or N 1 , 1 , 3 .

In what follows, we will prove the necessity and sufficiency parts of Theorems 9 and 10 in sections 2 and 3, respectively.

In this section, we construct sixteen classes of connected graphs which satisfy that δ ( G ) ≥ 2 and every odd branch bond has an edge branch, but they contain no even factor with exactly two components, see Figure 2.

Figure 2. Some classes of graphs without an even factor with exactly two components.

Proof of the necessity of Theorem 9. It is obviously that each of G 1 to G 16 in Figure 2 contains H as an induced subgraph and satisfies δ ( G ) ≥ 2 and every odd branch bond has an edge branch but it has no an even factor with exactly two components. Since the graphs G 1 , G 2 , …, G 16 have no common induced cycle, without loss of generality, we assume that H is an induced subgraph of G 7 , then H is K 1 , t ( t ≥ 3 ) or P i ( i ≥ 3 ) . Note that G 2 is K 1 , 3 -free, H is P i ( i ≥ 3 ) . Since the largest common induced path of G 1 to G 16 is P 5 , H is an induced subgraph of P 5 . This completes proof of the necessity of Theorem 9.

Proof of the necessity of Theorem 10. Let R , S be a pair of connected graphs of order at least three other than P 5 . It is obviously that each of G 1 to G 16 in Figure 2 satisfies δ ( G ) ≥ 2 and every odd branch bond has an edge branch but that it has no an even factor with exactly two components. Each of them contains at least one of R and S as an induced subgraph, without loss of generality, we assume that G 7 contains R as an induced subgraph. Then R is a tree or contains C 3 or C 4 . Now we distinguish the following two cases.

Case 1: R is a tree.

First suppose that R is a path. We note that the largest common induced path of G 5 , G 7 is P 5 , then R is an induced subgraph of P 5 if R is a path, which contradicts our assumption.

Next we suppose that R is K 1 , t ( i ≥ 3 ) . If R is K 1 , 4 ( i ≥ 4 ) , then we note that the graphs G 2 , G 3 , G 5 , G 6 , G 10 , G 11 , G 12 , G 14 , G 15 , G 16 are 2-edge-connected and K 1 , 4 -free, they contain S as an induced subgraph. Since the graphs G 2 , G 3 , G 5 , G 6 , G 10 , G 11 , G 12 , G 14 , G 15 , G 16 have no common cycle and G 2 is K 1 , 3 -free, S is a path. Note that the largest induced path of G 5 is P 5 , then S is an induced subgraph of P 5 , contradiction. Thus, R = K 1 , 3 . Since the graph G 2 is 2-edge-connected and K 1 , 3 -free, it contains S as an induced subgraph, thus S is a path or contains a cycle. Since the largest induced path of G 2 is P 8 , then S is an induced subgraph of P 8 . Next we consider S contains a cycle. Note that the maximal induced cycle of G 2 is triangle, S contains a triangle. Furthermore, S contains at most one triangle. Note that the maximal induced subgraph containing triangle in G 2 is Z 5 , B 3 , 2 or N 1 , 1 , 3 , S is an induced subgraph of Z 5 , B 3 , 2 or N 1 , 1 , 3 .

Finally suppose that R is a tree. Since the maximal induced subgraph of G 7 is T 3 , 1 , 1 , we have R = T 3 , 1 , 1 . Note that the graphs G 2 , G 5 , G 6 are 2-edge-connected and T 3 , 1 , 1 -free, they must contain S as a common induced subgraph. Since the graphs G 2 , G 5 , G 6 have no common induced cycle and G 2 is K 1 , 3 -free, S must be a path. Since the maximal induced path of G 5 is P 5 , then the maximal common induced subgraph of G 2 , G 5 , G 6 is P 5 , S is an induced subgraph of P 5 , contradiction. Then R = T 2 , 1 , 1 . Since the graph G 2 is 2-edge-connected and T 2 , 1 , 1 -free, it contains S as an induced subgraph, thus S is a path or contains a cycle. Since the largest induced path of G 2 is P 8 , then S is an induced subgraph of P 8 . Next we consider S contains a cycle. Note that the maximal induced cycle of G 2 is triangle, S contains a triangle. Furthermore, S contains at most one triangle. Note that the maximal induced subgraph containing triangle in G 2 is Z 5 , B 3 , 2 or N 1 , 1 , 3 , S is an induced subgraph of Z 5 , B 3 , 2 or N 1 , 1 , 3 . In fact, K 1 , 3 is an induced subgraph of T 2 , 1 , 1 , K 1 , 3 -free implies T 2 , 1 , 1 -free. Thus, { R , S } ≺ _ { T 2 , 1 , 1 , P 8 } , { T 2 , 1 , 1 , Z 5 } , { T 2 , 1 , 1 , B 3 , 2 } , { T 2 , 1 , 1 , N 3 , 1 , 1 } .

Case 2: R contains an induced C 4 or C 3 .

First assume that R contains an induced C 4 . Note that the graphs G 1 , G 2 , G 4 , G 10 , G 13 , G 14 are 2-edge-connected and C 4 -free, they must contain S as a common induced subgraph. Since the graphs G 1 , G 2 , G 4 , G 10 , G 13 , G 14 have no common induced cycle and G 2 is K 1 , 3 -free, S must be a path. Note that the maximal common induced path of G 1 , G 2 , G 4 , G 10 , G 13 , G 14 is P 8 , S is an induced subgraph of P 8 . On the other hand, assume S = P 8 . Since the graphs G 5 , G 6 , G 7 , G 8 , G 12 are P 8 -free and their maximal common induced subgraph containing C 4 is F 2 , R is an induced subgraph of F 2 . Thus, { R , S } ≺ _ { F 2 , P 8 } .

Finally suppose that R contains an induced C 3 . Note that the graphs G 1 , G 4 , G 5 , G 6 are 2-edge-connected and C 3 -free, they must contain S as an common induced subgraph. Since the graphs G 1 , G 4 , G 5 , G 6 have no common induced cycle, S is a tree. Since the maximal induced path of G 5 is P 5 , if S is a path, S is an induced subgraph of P 5 , contradiction. Then S is a tree. We note that the largest common induced subgraph the graphs G 1 , G 4 , G 5 , G 6 are K 1 , 3 or T 1 , 1 , 2 . We assume that S is T 1 , 1 , 2 . Note that the graph G 2 is T 1 , 1 , 2 -free, R is an induced subgraph G 2 . We note that the maximal induced subgraph containing a triangle in G 2 is Z 5 , B 3 , 2 or N 1 , 1 , 3 . Thus, { R , S } ≺ _ { T 2 , 1 , 1 , P 8 } , { T 2 , 1 , 1 , Z 5 } , { T 2 , 1 , 1 , B 3 , 2 } , { T 2 , 1 , 1 , N 3 , 1 , 1 } . This completes proof of the necessity of Theorem 10.

Before the start of this section, we still need some notations. Let G 1 and G 2 be two spanning subgraphs of G , and we can form the spanning subgraph G 1 Δ G 2 of G , whose vertex set is V ( G 1 ) ∪ V ( G 2 ) and edge set is E ( G 1 ) Δ E ( G 2 ) . This graph is called the symmetricdifference of G 1 and G 2 . Note that the symmetric difference of two even graphs remains an even graph. We denote the edge set of { u x , u y , v x , v y } by E ( { u , v } , { x , y } ) .

Let G be a connected graph admitting an even factor F with components Q 1 , Q 2 , ⋯ , Q t ( t ≥ 2 ) . Let Θ F ^ = { e ∈ E ( G ) \ E ( F ) } , where e is an edge with two end vertices in different components of F . Let C be an induced cycle of G with at least one edge of Θ F ^ . For any edge e = a b ∈ E ( C ) contained in a triangle, denoted by T ( a b ) and exactly one vertex in T ( a b ) that is not in C . Let

We also define an operation on the triangle T ( a b ) :

Then F C ( T ( a b ) ) is an even factor of G such that a , b are in the same components of F C ( T ( a b ) ) .

An edge of G is called a singular edge if it is not contained in any triangle. Otherwise, it is called nonsingular.

Lemma 11.Let G be a connected graph admitting an even factor. Let F be an even factor of G with components Q 1 , Q 2 , ⋯ , Q t ( t ≥ 2 ) such that it contains as few components as possible. Then for any edge u v ∈ Θ F ^ , u ∈ Q i , v ∈ Q j , it holds that:

1) u v is singular;

2) E ( N ( u ) ∩ V ( Q i ) , N ( v ) ∩ V ( Q j ) ) = ∅ .

Proof of Lemma 11.1) We choose an even factor F of G with components Q 1 , Q 2 , ⋯ , Q t ( t ≥ 2 ) such that it contains components as few as possible. Then u ∈ Q i , v ∈ Q j . Assume that u v is a nonsingular edge, u v is contained in a triangle u v w u . Then F Δ E ( u v w u ) is an even factor with fewer components than F , contradiction.

2) Conversely, we assume there exist two distinct vertices u 1 ∈ N ( u ) ∩ V ( Q i ) and v 1 ∈ N ( v ) ∩ V ( Q j ) such that u 1 v 1 ∈ E ( G ) . Then F Δ E ( u v v 1 u 1 u ) is an even factor with fewer components than F , contradiction. □

Proof of the sufficiency of Theorem 9. Let G be a { P 5 } -free graph. We choose an even factor F of G with components Q 1 , Q 2 , ⋯ , Q t ( t ≥ 2 ) such that it contains components as few as possible. Suppose that G has no even factor with exactly two components. Then the number of components t of any even factor satisfies t = 1 or t ≥ 3 . If t = 1 , then G = F is an even factor with exactly one component, so E \ F = ∅ . Then we may choose an induced cycle C = x x 1 ω 1 y 1 y x of G . Since G is an even subgraph, the degree of x is even. Thus, there exist x 2 ∈ N ( x ) \ { x 1 , y } and y 2 ∈ N ( y ) \ { y 1 , x } . However, G [ { x 2 , x , x 1 , ω 1 , y 1 } ] ≅ P 5 , a contradiction. Thus, t ≠ 1 , which implies t ≥ 2 .

To prove that t = 2 , we assume that t ≥ 3 . Then there is an edge x y ∈ Θ F ^ such that x ∈ Q i , y ∈ Q j and { i , j } ∈ { 1 , 2 , ⋯ , t } . Since G is 2-edge-connected, such an induced cycle C always exists. Then we choose an induced cycle C = x z z 1 ω 1 ⋯ ω l y 1 y x of G such that x y ∈ Θ F ^ ∩ E ( C ) and x z ∈ Θ F ^ ∩ E ( C ) . Let { x 1 } ⊆ N ( x ) ∩ V ( Q i ) , { y 1 } ⊆ N ( y ) ∩ V ( Q j ) and { z 1 } ⊆ N ( z ) ∩ V ( Q k ) . By Lemma 11, | V ( C ) | ≥ 5 . Since G is P 5 -free, | V ( C ) | ≤ 5 . Then | V ( C ) | = 5 . If ω 1 ∉ V ( Q i ) ∪ V ( Q j ) ∪ V ( Q k ) , then F = F Δ E ( C ) is an even factor with fewer components than F , such that Q i , Q j , Q k contains ω 1 in the same component, a contradiction. Then ω 1 ∈ V ( Q i ) ∪ V ( Q j ) ∪ V ( Q k ) . Without loss of generality, assume that ω 1 ∈ V ( Q k ) . By lemma 11, x y and x z are singular, we can obtain E ( { x 1 , x 2 } , { y 1 , y , y 2 } ) = ∅ , E ( { x } , { y 1 , y 2 } ) = ∅ , E ( { x 1 , x 2 } , { z 1 , z , z 2 } ) = ∅ and E ( { x } , { z 1 , z 2 } ) = ∅ , where { y 2 } ⊆ N ( y ) ∩ V ( Q j ) \ { y 1 } , { x 2 } ⊆ N ( x ) ∩ V ( Q i ) \ { x 1 } and { z 2 } ⊆ N ( z ) ∩ V ( Q k ) \ { z 1 } . By the choice of C , { x ω 1 , z ω 1 } ∩ E ( G ) = ∅ . We claim that z 1 z 2 ∉ E ( G ) . Otherwise, let T ( z ω 1 ) = z z 1 ω 1 z . If z ω 1 ∉ E ( F ) , then F C ( T ( z ω 1 ) ) = F Δ E ( x z z 1 ω 1 y x ) is an even factor of G such that Q i , Q j , Q k are in the same component. If z ω 1 ∈ E ( F ) and | { z z 1 , z ω 1 , z 1 ω 1 } ∩ E ( G ) | ≤ 2 , then F C ( T ( z ω 1 ) ) = F Δ E ( x z z 1 ω 1 y x ) Δ E ( z z 1 ω 1 z ) \ { z ω 1 } is an even factor of G such that Q i , Q j , Q k are in the same component. In this case, F C ( T ( z ω 1 ) ) has fewer components than F , a contradiction. However, G [ { y 1 , y , ω 1 , z 1 , z } ] ≅ P 5 , a contradiction. Then, without loss of generality, we assume that ω 1 ∈ V ( Q j ) . We claim that z 1 z 2 ∉ E ( G ) . Otherwise, let T ( y ω 1 ) = y y 1 ω 1 y . If y ω 1 ∉ E ( F ) , then F C ( T ( y 1 ω 1 ) ) = F Δ E ( x z z 1 ω 1 y x ) is an even factor of G such that Q i , Q j , Q k are in the same component. If y ω 1 ∈ E ( F ) and | { y y 1 , y ω 1 , y 1 ω 1 } ∩ E ( G ) | ≤ 2 , then F C ( T ( y ω 1 ) ) = F Δ E ( x z z 1 ω 1 y x ) Δ E ( y y 1 ω 1 y ) \ { y ω 1 } is an even factor of G such that Q i , Q j , Q k are in the same component. In this case, F C ( T ( y ω 1 ) ) has fewer components than F , a contradiction. However, G [ { y 1 , y , ω 1 , z 1 , z } ] ≅ P 5 , a contradiction. This completes proof of the sufficiency of Theorem 9. □

Proof of the sufficiency of Theorem 10. Let G be a 2-edge-connected graph. We may choose an even factor F of G with components Q 1 , Q 2 , ⋯ , Q t ( t ≥ 2 ) such that it contains components as few as possible. Suppose that G has no even factor with exactly two components. Then the number of components t of any even factor satisfies t = 1 or t ≥ 3 . Before present the proofs 1), 2), we note that a cycle always existis, since G is 2-edge-connected and C is an induced cycle. Then we can choose an induced cycle C = x z z 1 ω 1 ⋯ ω l y 1 y x , ( l ≥ 4 ) of G such that x y ∈ Θ F ^ ∩ E ( C ) , x z ∈ Θ F ^ ∩ E ( C ) and z 1 ∈ N C ( z ) ∩ V ( Q k ) and y 1 ∈ N C ( y ) ∩ V ( Q j ) . Next, we will prove parts 1) to 2) of Theorem 11, respectively.

1): Let G be a 2-edge-connected and { F 2 , P 8 } -free graph. Since G is P 8 -free, | V ( C ) | ≤ 8 . By lemma 11, | V ( C ) | ≥ 5 . If t = 1 , then G = F is an even factor with exactly one component, so E \ F = ∅ . Then we may choose an induced cycle C = x x 1 ω 1 y 1 y x of G . Since G is an even subgraph, the degree of x is even. Thus, there exist x 2 ∈ N ( x ) \ { x 1 , y } and y 2 ∈ N ( y ) \ { y 1 , x } . Since | V ( G ) | ≥ 10 , there exist at least three additional distinct vertices such that y 3 ∈ N ( y 1 ) \ { ω 1 , y } , y 4 ∈ N ( y 3 ) \ { y 1 } , y 5 ∈ N ( y 4 ) \ { y 3 } . Since G [ { x 2 , x , x 1 , ω 1 , y 1 , y 3 , y 4 , y 5 } ] ≇ P 8 , { ω 1 x 2 , y 2 y 1 } ⊆ E ( G ) . However, G [ { x 2 , ω 1 , x 1 , x , y , y 2 } ] ≅ F 2 , a contradiction. Thus, t ≠ 1 , which implies t ≥ 2 . Now we distinguish the following cases:

To prove that t = 2 , we assume that t ≥ 3 .

Case (1): | V ( C ) | = 5 .

If | V ( C ) | = 5 , then there are two edges x y ∈ Θ F ^ and x z ∈ Θ F ^ such that x ∈ Q i , y ∈ Q j , z ∈ Q k and { i , j , k } ∈ { 1 , 2 , ⋯ , t } . Let { x 1 , x 2 } ⊆ N ( x ) ∩ V ( Q i ) , { y 1 , y 2 } ⊆ N ( y ) ∩ V ( Q j ) and { z 1 , z 2 } ⊆ N ( z ) ∩ V ( Q k ) . By applying Lemma 11 to x y and x z , we can obtain E ( { x 1 , x 2 } , { y 1 , y , y 2 } ) = ∅ , E ( { x } , { y 1 , y 2 } ) = ∅ , E ( { x 1 , x 2 } , { z 1 , z , z 2 } ) = ∅ and E ( { x } , { z 1 , z 2 } ) = ∅ . Then we can choose an induced cycle C = x z z 1 ω 1 y x . Since Q i , Q j and Q k are different components in F , then at least one of { z 1 ω 1 , ω 1 y } is in Θ F ^ . If ω 1 ∉ V ( Q i ) ∪ V ( Q j ) ∪ V ( Q k ) , then F Δ E ( C ) is an even factor with fewer components than F such that Q i , Q j , Q k which contains ω 1 are all in the same component of F Δ E ( C ) , a contradiction. Then ω 1 ∈ V ( Q i ) ∪ V ( Q j ) ∪ V ( Q k ) , without loss of generality, we assume that ω 1 ∈ V ( Q k ) .

Claim 1: ω 1 is not adjacent to any vertex in ( N ( z ) ∩ V ( Q k ) ) \ { z 1 } .

Proof: By contradiction, we first assume that there exists a vertex z 0 ∈ ( N ( z ) ∩ V ( Q k ) ) \ { z 1 } such that z 0 ω 1 ∈ E ( G ) . Applying lemma 11(1) to x z , we have E ( { z 0 } , { x 1 , x 2 , x } ) = ∅ . Since C is an induced cycle, is { x ω 1 } ∩ E ( G ) = ∅ . Since G [ { ω 1 , z 1 , z , z 0 , x , y } ] ≇ F 2 , z 0 z 1 ∈ E ( G ) . Let T ( z z 1 ) = z z 1 z 0 z . If z z 1 ∉ E ( F ) , then F C ( T ( z z 1 ) ) = F Δ E ( x y ω 1 z 1 z x ) is an even factor of G such that Q i , Q j , Q k are in the same component. If z z 1 ∈ E ( F ) and | { z z 1 , z z 0 , z 1 z 0 } ∩ E ( G ) | ≤ 2 , then F C ( T ( z z 1 ) ) = F Δ E ( x y ω 1 z 1 z x ) ∪ E ( z z 1 z 0 z ) \ { z z 1 } is an even factor of G such that Q i , Q j , Q k are in the same component. In this case, F C ( T ( z z 1 ) ) has fewer components than F , which is a contradiction. This proves Claim 1. □

Since | V ( G ) | ≥ 10 , there exist additional distinct vertices such that z 3 ∈ N ( z 2 ) \ { z } , z 4 ∈ N ( z 3 ) \ { z 2 } , z 5 ∈ N ( z 4 ) \ { z 3 } such that E ( { z 3 } , { z , z 1 , ω 1 } ) = ∅ . First suppose that z 3 z 1 ∈ E ( G ) , then G [ { z 1 , z , z 2 , z 3 , z 4 , z 5 } ] ≅ F 2 , a contradiction. Next suppose that z 3 z ∈ E ( G ) . Let T ( z z 3 ) = z z 2 z 3 z . If z z 3 ∉ E ( F ) , then F C ( T ( z z 3 ) ) = F Δ E ( x y ω 1 z 1 z x ) is an even factor of G such that Q i , Q j , Q k are in the same component. If z z 3 ∈ E ( F ) and | { z z 3 , z z 2 , z 2 z 3 } ∩ E ( G ) | ≤ 2 , then F C ( T ( z z 3 ) ) = F Δ E ( x y ω 1 z 1 z x ) Δ E ( z z 2 z 3 z ) is an even factor of G such that Q i , Q j , Q k are in the same component. In this case, F C ( T ( z z 3 ) ) has fewer components than F , a contradiction. By claim 1, we have { z 3 ω 1 , z 4 ω 1 , z 5 ω 1 } ⊈ E ( G ) . Since G [ { z 4 , z 3 , z 2 , z , z 1 , ω 1 } ] ≇ F 2 , z z 4 ∉ E ( G ) . Since G [ { y 1 , y , x , z , z 2 , z 3 , z 4 , z 5 } ] ≇ P 8 , z z 5 ∈ E ( G ) . Since G [ { y 1 , y , x , z , z 2 , z 3 , z 4 , z 1 } ] ≇ P 8 , z 1 z 4 ∉ E ( G ) . We suppose that there exists a vertex { y 0 } ⊆ N ( y 1 ) ∩ V ( Q j ) \ { y 2 } . We claim that y 0 y ∉ E ( G ) . Otherwise, let T ( y y 0 ) = y y 0 y 1 y . If y y 0 ∉ E ( F ) , then F C ( T ( y y 0 ) ) = F Δ E ( x y ω 1 z 1 z x ) is an even factor of G such that Q i , Q j , Q k are in the same component. If y y 0 ∈ E ( F ) and | { y y 1 , y y 0 , y 1 y 0 } ∩ E ( G ) | ≤ 2 , then F C ( T ( y y 0 ) ) = F Δ E ( x y ω 1 z 1 z x ) Δ E ( y y 0 y 1 y ) is an even factor of G such that Q i , Q j , Q k are in the same component. In this case, F C ( T ( y y 0 ) ) has fewer components than F , a contradiction. Since G [ { y 0 , y 1 , y , x , z , z 2 , z 3 , z 4 } ] ≇ P 8 , z 2 ω 1 ∈ E ( G ) . However, G [ { z 2 , z , z 1 , ω 1 , y , y 1 } ] ≅ F 2 , a contradiction.

Then, without loss of generality, we assume that ω 1 ∈ V ( Q j ) . By the choice of C , we have z ω 1 ∉ E ( G ) . Since G [ { y 1 , y , ω 1 , z 1 , z , z 2 , z 3 , z 4 } ] ≇ P 8 , { y 1 ω 1 , z 3 z 1 } ⊆ E ( G ) . However, G [ { z 3 , z 2 , z , z 1 , ω 1 , y } ] ≅ F 2 , a contradiction.

Case(2): | V ( C ) | = 6 .

If | V ( C ) | = 6 , then we can choose an induced cycle C = x z z 1 ω 1 y 1 y x . Since Q i , Q j and Q k are different components in F , then at least one of { ω 1 z 1 , ω 1 y 1 } is in Θ F ^ . First suppose that ω 1 y 1 ∈ Θ F ^ . By applying Lemma 11 to ω 1 y 1 , we have { z 1 y 1 , z 1 y 0 , ω 1 y 0 , ω 1 y } ⊈ E ( G ) . Since G [ { y 0 , y 1 , ω 1 , z 1 , z , z 2 , z 3 , z 4 } ] ≇ P 8 , z 2 ω 1 ∈ E ( G ) . However, G [ { z 2 , z , z 1 , ω 1 , y 1 , y 0 } ] ≅ F 2 , a contradiction. Then ω 1 z 1 ∈ Θ F ^ . By applying Lemma 11 to ω 1 z 1 , we have { z y 1 , z ω 1 } ⊈ E ( G ) . Since G [ { z 3 , z 2 , z , z 1 , ω 1 , y 1 , y , y 2 } ] ≇ P 8 , z 3 z 1 ∈ E ( G ) . However, G [ { z 3 , z 2 , z , z 1 , ω 1 , y 1 } ] ≅ F 2 , a contradiction.