Fractional Factors, Component Factors and Isolated Vertex Conditions in Graphs

For a graph $G = (V, E)$, a {\em fractional $[a, b]$-factor} is a real valued function $h:E(G)\to [0,1]$ that satisfies $a \le ~ \sum_{e\in E_G(v)} h(e) ~ \le b$ for all $ v\in V(G)$, where $a$ and $b$ are real numbers and $E_G(v)$ denotes the set of edges incident with $v$. In this paper, we prove that the condition $\mathit{iso}(G-S) \le (k+\frac{1}{2})|S|$ is equivalent to the existence of fractional $[1,k+ \frac{1}{2}]$-factors, where ${\mathit{iso}}(G-S)$ denotes the number of isolated vertices in $G-S$. Using fractional factors as a tool, we construct component factors under the given isolated conditions. Namely, (i) a graph $G$ has a $\{P_2,C_3,P_5, \mathcal{T}(3)\}$-factor if and only if $\mathit{iso}(G-S) \le \frac{3}{2}|S|$ for all $S\subset V(G)$; (ii) a graph $G$ has a $\{K_{1,1}, K_{1,2}, \ldots,$ $K_{1,k}, \mathcal{T}(2k+1)\}$-factor ($k\ge 2$) if and only if $\mathit{iso}(G-S) \le (k+\frac{1}{2})|S|$ for all $S\subset V(G)$, where $\mathcal{T}(3)$ and $\mathcal{T}(2k+1)$ are two special families of trees.


Introduction
In this paper, we mainly consider finite simple graphs, which have neither loops nor multiple edges. A graph that has multiple edges but has no loops is referred as a multigraph. When defining notation and definitions, we often referred a multigraph as a graph for convenience. Let G be a graph with vertex set V (G) and edge set E(G). The number of vertices of G is called its order and denoted by |G|. On the other hand, the number of edges in G is called its size and denoted by ||G||.
For a set X, the cardinality of X is denoted by |X| or #X. For a vertex v of a graph G, the degree of v in G is denoted by deg G (v). For two vertices x and y of G, an edge joining them is denoted by xy or yx. We denote by Iso(G) the set of isolated vertices of G, and by iso(G) the number of isolated vertices in G. Thus iso(G − S) = |Iso(G)|. For two disjoint vertex sets X and Y of G, the set of edges of G joining X to Y is written as E G (X, Y ) and e G (X, Y ) := |E G (X, Y )|. P n and C n are the path and the cycle of order n, respectively. The set of non-negative integers is denoted by Z * = {0} ∪ Z + .
Let G be a graph, and g, f : V (G) → Z * be two integer-valued functions with g ≤ f , that is, 0 ≤ g(x) ≤ f (x) for all x ∈ V (G). Then a spanning subgraph F of G is called a (g, f )-factor of G if g(x) ≤ deg F (x) ≤ f (x) for all x ∈ V (G). For a vertex v of G, let E G (v) denote the set of edges of G incident with v. For real-valued functions g, f : V (G) → R with g ≤ f , a fractional (g, f )-factor is a function h : E(G) → [0, 1] that satisfies the following condition: where deg h (v) is called the h-degree of v and h(e) is a real number between 0 and 1 including 0 and 1. If the values of h are 0 and 1 only, then a fractional (g, f )-factor becomes a (g, f )-factor. Many results on fractional factors of graphs can be found in [9].
To study fractional factors, Yang, Ma and Liu [8] introduced a new parameter, isolated toughness of a graph G, denoted by I(G), which is defined as For a set S of connected graphs, a spanning subgraph F of G is called an S-factor if each component of F is isomorphic to an element of S (see (1) of Figure 1). For a set S of positive integers, a spanning subgraph F of G is called an S-factor of G if deg F (x) ∈ S for all vertices x of G. For an integer k ≥ 0, the set of vertices of G with degree k is denoted by V k (G), namely, For a tree T , the set of leaves is denoted by Leaf (T ), i.e., V 1 (T ). An edge of T incident with a leaf is called a pendant edge. In particular, the number of leaves of T is equal to that of pendant edges of T .
We define a special class of trees T (3): for any {1, 3}-tree R (i.e., every vertex has degree 1 or 3), a new tree T R is obtained from R by inserting a new vertex of degree 2 into every edge of R, and by adding a new pendant edge together its endpoint to every leaf of R (see (2) and (3) of Figure 1). Then the tree T R is a {1, 2, 3}-tree having ||R|| + |Leaf (R)| vertices of degree 2 and has the same number of leaves as R. Also there is one-to-one correspondence between V 3 (R) and V 3 (T R ). The collection of such {1, 2, 3}-trees T R generated from all {1, 3}-trees R is denoted by T (3). A more general class of trees, T (2k + 1) (k ≥ 2), will be defined in Section 2. Tutte [7] established a relationship between isolated 1-tough graphs and {K 2 , C n : n ≥ 3}-factors.
Amahashi and Kano [2] extended Theorem 1 and gave a characterization for isolated 1/k-tough graphs in term of star factors.
Theorem 2 (Amahashi and Kano [2]) Let k ≥ 2 be an integer. A graph G has a {K 1,j : 1 ≤ j ≤ k}-factor if and only if Kano, Lu and Yu [5] obtained a sufficient condition for isolated 2-tough graphs to have a components factor.
Theorem 3 (Kano, Lu and Yu [5] Kano and Saito [6] as well as Zhang, Yan and Kano [10] used isolated k-toughness to ensure the existence of special classes of component factors.

Theorem 4 (Kano and Saito
Theorem 5 (Zhang, Yan and Kano [10]) Let k ≥ 2 be an integer. A graph G has a {K 1,j , K 2k : In this paper, we carry on the investigations along the same direction mentioned above and obtain the factor characterizations of I(G) = 2 3 and I(G) = 2 2k+1 (k ≥ 2): Theorem 7 Let k ≥ 2 be an integer. Then a graph G has a {K 1,1 , K 1,2 , . . . , 2 Proofs of Theorems 6 and 7 For a function f : V (G) → Z * and a vertex set X of G, we write The tools for proving Theorems 6 and 7 are fractional factors. We first characterize the condition (3) in fractional [1, k + 1 2 ]-factors (Theorem 9), and then show that the minimal fractional factors are the desired component factors. In establishing the link between the condition (3) and fractional factors, we need the following theorem.
Theorem 8 (Heinrich et al. [4], Anstee [3]) Let G be a multigraph and where Theorem 9 Let k ≥ 1 be an integer and G be a graph. Then G has a fractional [1, k Proof. Assume that G satisfies (5). Let G * denote the multigraph obtained from G by replacing each edge e of G by two parallel edges e(1) and e (2).
Then g < f , and for every S ⊂ V (G * ), we have Thus it follows from the above equality and (5) that Hence by Theorem 8, G * has a (g, f )-factor F . Now we construct a fractional Hence iso(G − S) ≤ (k + 1 2 )|S|, i.e., (5) holds. 2 Proof of Theorem 6. We first show that every tree T ∈ T (3) satisfies the condition (2). Define a function h : E(T ) → { 1 2 , 1} as follows: for every pendant edge e 1 of T , let h(e 1 ) = 1 and for any other edge e 2 , let h(e 2 ) = 1 2 . Since T is a {1, 2, 3}-tree and no pendant edge is incident with a vertex of degree 3, h is a fractional [1, 3 2 ]-factor. Hence, by Theorem 9, T satisfies the condition (2).
Assume that G has a {P 2 , C 3 , Hence the necessity is proved.
Next we prove the sufficiency. By Theorem 9, G has a fractional [1, 3 2 ]factor h with values in {0, 1 2 , 1}. We call an edge e 1 with h(e 1 ) = 1 a red edge and an edge e 2 with h(e 2 ) = 1 2 a blue edge. Let F be the subgraph of G induced by the set of all red and blue edges. Namely, F is obtained from G by removing all the edges e 3 with h(e 3 ) = 0. Since h is a fractional [1, 3 2 ]-factor, F is a spanning subgraph of G, and for every vertex v of G, one of the following two statements holds: (i) no red edge is incident with v and two or three blue edges are incident with v; or (ii) exactly one red edge is incident with v and at most one blue edge is incident with v.
Choose a fractional [1, 3 2 ]-factor h so that the number of edges in F is as small as possible. For a convenience, we also call F a fractional [1, 3 2 ]-factor. For every vertex v, we call the number of edges of F incident with v the degree of v and denote it by deg F (v). It is clear that deg Proof. Assume that F contains a cycle C. First assume that C is of even order. Take a perfect matching M of C, and recolor all the edges of M red, and remove all the edges in C − M . Then the resulting subgraph is a new fractional [1, 3 2 ]-factor with red and blue edges, but its size is smaller than F , which contradicts the choice of F . Hence C is of odd order.
Assume that C has two adjacent vertices u 1 and u 2 with degree 3 in F . Then F −u 1 u 2 is a new fractional [1, 3 2 ]-factor with fewer edges than F , which contradicts to the choice of F . Hence if a vertex v of C has degree 3 in F , then the two neighbors of v in C have degree 2 in F . Assume that C has a vertex v with deg F (v) = 3. Let u 1 and u 2 be the two neighbors of v in C. Take a perfect matching M of C − v. Recolor the edges of M red, and remove all the edges of (C − v) − M and vu 2 . Since vu 1 and vu 2 are both blue edges, we obtain a new fractional [1, 3 2 ]-factor with fewer edges than F , which is a contradiction. Hence C is a component of F .
Moreover, it is easy to see that an odd cycle of C order at least 5 has a {P 2 , P 5 }-factor F C . Remove all the edges of C not contained in F C , recolor the edges contained in P 2 of F C red, and two pendant edges of P 5 of F C red and the remaining two edges of P 5 of F C blue. Then we obtain a new fractional [1, 3 2 ]-factor with fewer edges than F , a contradiction. Therefore every cycle contained in F is C 3 . Consequently Claim 1 is proved.
For the simplicity of statements, from now on, we will use "another fractional [1, 3 2 ]-factor" to replace the phase "a new fractional [1, 3 2 ]-factor with fewer edges than F ".
Claim 2. Every non-cycle component of F is P 2 , P 5 or a tree of T (3).
Proof. Let x and y be two vertices of degree 3 in F such that they are adjacent or connected by a path whose all inner vertices have degree 2 in F . If x and y are adjacent in F , then F − xy is another fractional [1, 3 2 ]-factor, a contradiction. Assume that x and y are connected by a path (x, u 1 , u 2 , . . . , u n , y) of length at least 3 (i.e., n ≥ 2) such that every u i has degree 2 in F . Then remove u n y, recolor u n−1 u n red, and recolor all remaining edges of the path blue. Then resulting subgraph is another fractional [1, 3 2 ]-factor, a contradiction. Therefore, (a) if two vertices of F with degree 3 are connected by a path in F whose inner vertices have degree 2 in F , then the length of the path is 2.
Let z be a leaf of F and x be a vertex of degree 3 in F . If z and x are adjacent, then the edge xz is red and so deg h (x) ≥ 2, which is impossible. Hence z and x are not adjacent. Assume that z and x are connected by a path (z, u 1 , u 2 , . . . , u n , x) with deg F (u i ) = 2 for every i. Then zu 1 is red. First assume that n ≥ 3. Then remove u n x, recolor u n−1 u n red, and recolor all the remaining edges of the path except zu 1 blue. Then the resulting subgraph is another fractional [1, 3 2 ]-factor, a contradiction. Next assume n = 1. Then by removing u 1 x, we obtain another fractional [1, 3 2 ]-factor, a contradiction again. Therefore, Consequently, if a component D of F contains at least two vertices of degree 3, then by (a) and (b), D is a tree of T (3). If D has exactly one vertex of degree 3, then by the property (b), D is also a tree of T (3). If D has no vertex of degree 3, then D is a path. It is obvious that P 3 has no fractional [1, 3 2 ]-factor, and so D is not P 3 . If D is a path of even order, then D has a P 2 -factor, and so it contradicts to the minimality of F . If D is a path P n of odd order with n ≥ 7, then D has a {P 2 , P 5 }-factor and thus contradicts to the minimality of F . Hence Claim 2 holds.
Clearly, Claims 1 and 2 imply the sufficiency.
2 To state Theorem 7, we need a new class of trees T (2k + 1). Let k ≥ 2 be an integer and R be a tree that satisfies the following conditions: for every vertex v ∈ V (R),  (1) and (2) of Figure 2). For such a tree R, we obtain a new tree T R as follows: (iii) insert a new vertex of degree 2 into each edge of R − Leaf (R), and  Figure 2). The set of such trees T R for all trees R satisfying (6) is denoted by T (2k + 1). Note that the construction of T (3) and that of T (2k + 1) with k ≥ 2 are similar, but adding pendant edges to some vertices of R − Leaf (R) is not defined in the construction of T (3).
Proof of Theorem 7. For any tree T ∈ T (2k + 1), since T has a fractional [1, k + 1 2 ]-factor h of values { 1 2 , 1}, T satisfies (3) by Theorem 9. Assume that G has a {K 1,1 , K 1,2 , . . . , K 1,k , T (2k + 1)}-factor F . Let D 1 , D 2 , . . . , D m be the components of F . Then each D i is K 1,s for some 1 ≤ s ≤ k, or a tree in T (2k + 1). Thus iso( Hence the necessity is proved. Next we prove the sufficiency. Assume that G satisfies (3). By Theorem 9, G has a fractional [1, k + 1 2 ]-factor h with values in {0, 1 2 , 1}. We call an edge e 1 with h(e 1 ) = 1 a red edge and an edge e 2 with h(e 2 ) = 1 2 a blue edge. Let F be the subgraph of G induced by the set of all red and blue edges. Namely, F is obtained from G by removing all the edges e 3 with h(e 3 ) = 0, and since h is a fractional [1, k + 1 2 ]-factor, F is a spanning subgraph. Choose a fractional [1, k + 1 2 ]-factor h of G so that the number of edges in F is as small as possible.
For a vertex v, we call the number of edges of F incident with v the degree of v in F and denote it by deg F (v). It is clear that 1 ≤ deg F (v) ≤ 2k + 1. On the other hand, the h-degree of v can be expressed as deg h (v) = 1 + 1 2 t for some integer t, 0 ≤ t ≤ 2k − 1.
Claim 1. F contains no cycle, i.e., F is a forest.
Proof. Suppose that F contains a cycle C. First assume that C is of even order. Take a perfect matching M of C, and recolor all edges of M red, and remove all edges in C − M . Then the resulting subgraph is a new fractional [1, k + 1 2 ]-factor of G, but its size is smaller than F , a contradiction. Hence C is of odd order.
As in the proof of Theorem 6, we will replace "a new fractional [1, k + 1 2 ]factor with less edges than F " by "another fractional factor" for the simplicity in the rest of the proof.
Assume that C has two adjacent vertices v 1 and v 2 with degree at least 3 in F . Then F − v 1 v 2 is another fractional factor, a contradiction. Hence if a vertex v of C has degree at least 3 in F , then the two neighbors of v in C have degree 2 in F . Assume that C has a vertex v with deg F (v) ≥ 3. Let u 1 and u 2 be the two neighbors of v in C. Take a perfect matching M of C − v, and recolor all edges of M red, and remove all edges of (C − v) − M and the edge vu 2 . Then we obtain another fractional factor, a contradiction. Hence C is a component of F .
It is easy to see that C has a {P 2 = K 1,1 , P 3 = K 1,2 }-factor F C . We recolor all edges of P 2 -components and P 3 -components of F C red, and remove all other edges of C. Then we obtain another fractional factor, a contradiction. Therefore F has no cycles, and the claim is proved. Claim 2. Let x and y be two vertices of degree at least 3 in F . Then x and y are not adjacent in F . If x and y are connected by a path whose all inner vertices have degree 2 in F , then the length of the path is 2, and deg h (x) = deg h (y) = k + 1 2 and the two edges in the path are blue edges. Proof. If x and y are adjacent in F , then F − xy is another fractional factor, a contradiction. Assume that x and y are connected by a path (x, u 1 , u 2 , . . . , u n , y) with deg F (u i ) = 2 (1 ≤ i ≤ n). If n ≥ 2, then by removing u n y and recoloring u n−1 u n red and all the remaining edges of the path blue, the resulting subgraph is another fractional factor, a contradiction. Therefore n = 1, and the path is (x, u 1 , y).
If deg h (x) < k + 1 2 , then by removing u 1 y and recoloring xu 1 red, we obtain another fractional factor, a contradiction. Hence deg h (x) = deg h (y) = k + 1 2 by the symmetry. If xu 1 is red, then removing u 1 y we obtain another fractional factor, a contradiction. Therefore xu 1 and yu 1 are blue edges, and the claim holds.
Claim 3. A leaf z in F is either contained in a star component, or adjacent to a vertex x with deg F (x) ≥ 3 and deg h (x) = k + 1 2 . Proof. Let z be a leaf of F , and D be the component of F containing z. Assume that D is a path (z, u 1 , u 2 , . . . , u n , y) such that deg F (u i ) = 2 (1 ≤ i ≤ n) and deg F (y) = 1. If n ≥ 2, then by removing u 1 u 2 and recoloring u 2 u 3 red, we obtain another fractional factor, a contradiction. If n = 1, then D = P 3 = K 1,2 , which is a star.
Next assume that F contains a path (z, u 1 , u 2 , . . . , u n , x) such that deg F (u i ) = 2 (1 ≤ i ≤ n) and x has degree at least 3 in F . If n ≥ 2, then by removing u n x, and recoloring u n−1 u n and zu 1 red, and recoloring all other remaining edges (if any) of the path blue, we obtain another fractional factor, a contradiction. If n = 1, then F − u 1 x is another fractional [1, k + 1 2 ]-factor, a contradiction. Therefore z and x are adjacent.
Moreover, if D contains exactly one vertex x of degree at least 3, by the same argument given above, we see that every leaf of D is adjacent to x and thus D is a star; otherwise, D contains another vertex y of degree at least 3, then by Claim 2, deg h (x) = k + 1 2 . Thus the claim is proved. Claim 4. (i) If uv is an edge such that deg F (u) ≥ 3 and deg F (v) = 2, then deg h (u) = k + 1 2 and uv is a blue edge. (ii) If xy is a red edge, then one of x and y is a leaf of F . Proof. Let uv be an edge of F such that deg F (u) ≥ 3, deg F (v) = 2 and deg h (u) < k + 1 2 . Let z 1 be a vertex adjacent to v. If deg F (z 1 ) ≥ 3, then by Claim 2, we have deg h (u) = k + 1 2 , a contradiction. If deg F (z 1 ) = 1, then it contradicts to Claim 3. Hence deg F (z 1 ) = 2. By removing uv and recoloring vz 1 red, we obtain another fractional factor, a contradiction. Hence if uv is an edge with deg F (u) ≥ 3 and deg F (v) = 2, then deg h (u) = k + 1 2 . If the statement (ii) is true, then uv is a blue edge, and so in order to show that uv is a blue edge, it suffices to prove (ii).