Isolated toughness and path-factor uniform graphs (II)

Abstract

A spanning subgraph F of G is called a path-factor if each component of F is a path. A $P_{\ge k}$-factor of G means a path-factor such that each component is a path with at least k vertices, where $k\ge 2$ is an integer. A graph G is called a $P_{\ge k}$-factor covered graph if for each $e\in E(G)$, G has a $P_{\ge k}$-factor covering e. A graph G is called a $P_{\ge k}$-factor uniform graph if for any two different edges $e_1,e_2\in E(G)$, G has a $P_{\ge k}$-factor covering $e_1$ and avoiding $e_2$. In other word, a graph G is called a $P_{\ge k}$-factor uniform graph if for any $e\in E(G)$, the graph $G-e$ is a $P_{\ge k}$-factor covered graph. In this article, we demonstrate that (i) an $(r+3)$-edge-connected graph G is a $P_{\ge 2}$-factor uniform graph if its isolated toughness $I(G)>\frac{r+3}{2r+3}$, where r is a nonnegative integer; (ii) an $(r+3)$-edge-connected graph G is a $P_{\ge 3}$-factor uniform graph if its isolated toughness $I(G)>\frac{3r+6}{2r+3}$, where r is a nonnegative integer. Furthermore, we claim that these conditions on isolated toughness and edge-connectivity in our main results are best possible in some sense.

1 Introduction

The graphs discussed here are finite, undirected and simple. Let G be a graph. The vertex set and the edge set of G are denoted by V(G) and E(G), respectively. Denote by $d_G(v)$ the degree of a vertex v in G. Let i(G) and $\omega (G)$ denote the number of isolated vertices and connected components in G, respectively. For any $X\subseteq V(G)$, G[X] is the subgraph of G induced on X, and $G-X$ denotes the subgraph $G[V(G)\setminus X]$. For any $E'\subseteq E(G)$, $G-E'$ denotes the subgraph derived from G by removing $E'$. A vertex subset X of G is called an independent set if G[X] has no edges. The path and the complete graph with n vertices are denoted by $P_n$ and $K_n$, respectively. We use $G_1\vee G_2$ to denote the join of two disjoint graphs $G_1$ and $G_2$, and $G_1\cup G_2$ to denote the union of two disjoint graphs $G_1$ and $G_2$.

Yang, Ma and Liu [14] introduced the notion of isolated toughness, which is defined by

$$\begin{aligned} I(G)=\min \{\frac{|X|}{i(G-X)}:X\subseteq V(G), i(G-X)\ge 2\} \end{aligned}$$

if G is not a complete graph; otherwise, $I(G)=+\infty$.

A spanning subgraph F of G is called a path-factor if each component of F is a path. A $P_{\ge k}$-factor of G means a path-factor such that each component is a path with at least k vertices, where $k\ge 2$ is an integer.

Johnson et al. [7] studied the existence of path-factors in graphs. Egawa and Furuya [3] posed some sufficient conditions for a graph to admit a path-factor. Asratian and Casselgren [2] derived a sufficient condition for the existence of path-factors in graphs. Ando et al. [1] verified that a claw-free graph with minimum degree at least k admits a $P_{\ge k+1}$-factor. Kano, Lee and Suzuki [10] proved that every connected cubic bipartite graph with at least 8 vertices admits a $P_{\ge 8}$-factor. Johansson [6] gave an El-Zahár type condition ensuring path-factors in graphs. Kano, Lu and Yu [11] claimed that a graph G with $i(G-X)\le \frac{2}{3}|X|$ for any $X\subseteq V(G)$ admits a $P_{\ge 3}$-factor. Zhou [17,18,19], Zhou, Bian and Pan [20], Zhou, Sun and Liu [23], Zhou, Wu and Bian [24], Zhou, Wu and Xu [25],and Gao, Wang and Chen [5] presented some sufficient conditions for graphs admitting $P_{\ge 3}$-factors with given properties. Some other results on graph factors see [13, 16, 21]. Las Vergnas [12] showed a criterion for a graph with a $P_{\ge 2}$-factor.

Theorem 1

( [12]). A graph G has a $P_{\ge 2}$-factor if and only if

$$\begin{aligned} i(G-X)\le 2|X| \end{aligned}$$

for any $X\subseteq V(G)$.

A graph R is called a factor-critical graph if for any $v\in V(R)$, $R-v$ has a 1-factor. A graph H is called a sun if $H=K_1$, $H=K_2$ or H is the corona of a factor-critical graph R with at least three vertices, namely, H is derived from R by adding a new vertex $u=u(v)$ together with a new edge vu for each $v\in V(R)$. A sun with at least six vertices is called a big sun. The number of sun components of G is denoted by sun(G). Kaneko [8] presented a characterization of a graph with a $P_{\ge 3}$-factor, and Kano, Katona and Király [9] gave a shorter proof.

Theorem 2

( [8, 9]). A graph G has a $P_{\ge 3}$-factor if and only if

$$\begin{aligned} sun(G-X)\le 2|X| \end{aligned}$$

for any $X\subseteq V(G)$.

A graph G is called a $P_{\ge k}$-factor covered graph if for each $e\in E(G)$, G has a $P_{\ge k}$-factor covering e, which was first defined by Zhang and Zhou [15]. Furthermore, they put forward two characterizations for a graph to a $P_{\ge 2}$-factor covered graph and $P_{\ge 3}$-factor covered graph, which are stated as follows.

Theorem 3

( [15]). A connected graph G is a $P_{\ge 2}$-factor covered graph if and only if

$$\begin{aligned} i(G-X)\le 2|X|-\varepsilon _1(X) \end{aligned}$$

for any $X\subseteq V(G)$, where $\varepsilon _1(X)$ is defined by

$$\begin{aligned} \varepsilon _1(X)=\left\{ \begin{array}{ll} 2,&{}if \ X \ is \ not \ an \ independent \ set;\\ 1,&{}if \ X \ is \ a \ nonempty \ independent \ set, \ and \ G-X \ admits\\ &{}a \ nontrivial \ component;\\ 0,&{}otherwise.\\ \end{array} \right. \end{aligned}$$

Theorem 4

( [15]). A connected graph G is a $P_{\ge 3}$-factor covered graph if and only if

$$\begin{aligned} sun(G-X)\le 2|X|-\varepsilon _2(X) \end{aligned}$$

for any subset X of V(G), where $\varepsilon _2(X)$ is defined by

$$\begin{aligned} \varepsilon _2(X)=\left\{ \begin{array}{ll} 2,&{}if \ X \ is \ not \ an \ independent \ set;\\ 1,&{}if \ X \ is \ a \ nonempty \ independent \ set, \ and \ G-X \ admits\\ &{}a \ non-sun \ component;\\ 0,&{}otherwise.\\ \end{array} \right. \end{aligned}$$

A graph G is called a $P_{\ge k}$-factor uniform graph if for any two different edges $e_1,e_2\in E(G)$, G has a $P_{\ge k}$-factor covering $e_1$ and avoiding $e_2$. In other word, a graph G is called a $P_{\ge k}$-factor uniform graph if for any $e\in E(G)$, the graph $G-e$ is a $P_{\ge k}$-factor covered graph. Gao and Wang [4] derived a result on the existence of $P_{\ge 3}$-factor uniform graphs. Zhou, Sun and Liu [22] posed two isolated toughness conditions for a graph to be a $P_{\ge 2}$-factor uniform graph and $P_{\ge 3}$-factor uniform graph, which are stated as follows.

Theorem 5

( [22]). A 3-edge-connected graph G is a $P_{\ge 2}$-factor uniform graph if its isolated toughness $I(G)>1$.

Theorem 6

( [22]). A 3-edge-connected graph G is a $P_{\ge 3}$-factor uniform graph if its isolated toughness $I(G)>2$.

In Theorems 5 and 6, the conditions on I(G) are sharp. However, it is natural to expect that we can weaken the condition on I(G) if we replace the assumption that G is 3-edge-connected by a stronger assumption. Along this line, we derive the following results.

Theorem 7

Let r be a nonnegative integer. An $(r+3)$-edge-connected graph G is a $P_{\ge 2}$-factor uniform graph if its isolated toughness $I(G)>\frac{r+3}{2r+3}$.

Theorem 8

Let r be a nonnegative integer. An $(r+3)$-edge-connected graph G is a $P_{\ge 3}$-factor uniform graph if its isolated toughness $I(G)>\frac{3r+6}{2r+3}$.

2 The proof of Theorem 7

Proof of Theorem 7

Theorem 7 holds obviously for a complete graph. Hence, we may assume that G is not a complete graph.

We proceed by contradiction. Assume that there exists an edge $e=uv$ in G such that $G'=G-e$ is not a $P_{\ge 2}$-factor covered graph. It follows from Theorem 3 that

$$\begin{aligned} i(G'-X)\ge 2|X|-\varepsilon _1(X)+1 \end{aligned}$$

(1)

for some vertex subset X of $G'$.

The following proof will be divided into three cases by the value of |X|.

Case 1.$0\le |X|\le r+1$.

According to (1) and $\varepsilon _1(X)\le |X|$, we obtain

$$\begin{aligned} i(G'-X)\ge 2|X|-\varepsilon _1(X)+1\ge |X|+1, \end{aligned}$$

which implies that $G'-X$ has at least one isolated vertex w, and so $d_{G'-X}(w)=0$. Thus, we have

$$\begin{aligned} d_G(w)\le d_{G'}(w)+1\le d_{G'-X}(w)+|X|+1=|X|+1\le r+2, \end{aligned}$$

which contradicts that G is $(r+3)$-edge-connected.

Case 2.$|X|=r+2$.

In terms of (1) and $\varepsilon _1(X)\le 2$, we admit

$$\begin{aligned} i(G'-X)\ge 2|X|-\varepsilon _1(X)+1\ge 2|X|-1. \end{aligned}$$

(2)

Note that $i(G-X)\ge i(G'-X)-2$. Combining this with (2), we get

$$\begin{aligned} i(G-X)\ge i(G'-X)-2\ge 2|X|-3=2(r+2)-3=2r+1, \end{aligned}$$

which implies that there exists $w\in V(G-X)$ with $d_{G-X}(w)=0$. Since G is $(r+3)$-edge-connected, we have

$$\begin{aligned} r+3\le d_G(w)\le d_{G-X}(w)+|X|=|X|=r+2, \end{aligned}$$

which is a contradiction.

Case 3.$|X|\ge r+3$.

Note that $i(G-X)\ge i(G'-X)-2$. It follows from (1) and $\varepsilon _1(X)\le 2$ that

$$\begin{aligned} i(G-X)\ge i(G'-X)-2\ge 2|X|-\varepsilon _1(X)+1-2\ge 2|X|-3\ge 2r+3. \end{aligned}$$

(3)

In light of (3) and the definition of I(G), we derive

$$\begin{aligned} I(G)\le & {} \frac{|X|}{i(G-X)}\le \frac{|X|}{2|X|-3}=\frac{1}{2}+\frac{3}{4|X|-6}\\\le & {} \frac{1}{2}+\frac{3}{4(r+3)-6}=\frac{1}{2}+\frac{3}{4r+6}\\= & {} \frac{r+3}{2r+3}, \end{aligned}$$

which contradicts $I(G)>\frac{r+3}{2r+3}$. The proof of Theorem 7 is complete. $\square$

Remark 1

We now show that $I(G)>\frac{r+3}{2r+3}$ in Theorem 7 cannot be weakened to $I(G)\ge \frac{r+3}{2r+3}$.

We construct a graph $G=K_{r+3}\vee ((2r+3)K_1)\cup K_2$, where r is a nonnegative integer. Then $I(G)=\frac{|V(K_{r+3})|}{i(G-V(K_{r+3}))} =\frac{r+3}{2r+3}$ and G is $(r+3)$-edge-connected. Write $G'=G-e$ for $e\in E(K_2)$. Let $X=V(K_{r+3})$. Then $|X|=r+3$ and $\varepsilon _1(X)=2$. Thus, we admit

$$\begin{aligned} i(G'-X)=2r+5>2r+4=2|X|-\varepsilon _1(X). \end{aligned}$$

In light of Theorem 3, $G'$ is not a $P_{\ge 2}$-factor covered graph, and so G is not a $P_{\ge 2}$-factor uniform graph.

Remark 2

The condition that G is $(r+3)$-edge-connected in Theorem 7 cannot be replaced by G being $(r+2)$-edge-connected.

To show this, we construct a graph $G=K_{r+2}\vee ((2r+1)K_1\cup K_2)$, where $r\ge 1$ is an integer. Then G is $(r+2)$-edge-connected and $I(G)=\frac{|V(K_{r+2})|}{i(G-V(K_{r+2}))}=\frac{r+2}{2r+1}>\frac{r+3}{2r+3}$. Let $G'=G-e$ for $e\in E(K_2)$ and $X=V(K_{r+2})$. Then $\varepsilon _1(X)=2$, and so

$$\begin{aligned} i(G'-X)=2r+3>2r+2=2|X|-\varepsilon _1(X). \end{aligned}$$

According to Theorem 3, $G'$ is not a $P_{\ge 2}$-factor covered graph, and so G is not a $P_{\ge 2}$-factor uniform graph.

3 The proof of Theorem 8

Proof of Theorem 8

Theorem 8 is true for a complete graph. Therefore, we may assume that G is not a complete graph.

Theorem 8 holds for $r=0$ by Theorem 6. In what follows, we may assume $r\ge 1$. We proceed by contradiction. Assume that there exists an edge $e=uv$ in G such that $G'=G-e$ is not a $P_{\ge 3}$-factor covered graph. Then by Theorem 4 we obtain

$$\begin{aligned} sun(G'-X)\ge 2|X|-\varepsilon _2(X)+1 \end{aligned}$$

(4)

for some vertex subset X of $G'$.

Claim 1. $|X|\ge r+2$.

Proof

If $0\le |X|\le r$, then it follows from (4) and $\varepsilon _2(X)\le |X|$ that

$$\begin{aligned} sun(G'-X)\ge 2|X|-\varepsilon _2(X)+1\ge |X|+1\ge 1, \end{aligned}$$

which implies that there exists $w\in V(G'-X)$ such that $d_{G'-X}(w)\le 1$. Thus, we derive

$$\begin{aligned} d_G(w)\le d_{G'}(w)+1\le d_{G'-X}(w)+|X|+1\le |X|+2\le r+2, \end{aligned}$$

which contradicts that G is $(r+3)$-edge-connected.

If $|X|=r+1$, the by (4), $\varepsilon _2(X)\le 2$ and $r\ge 1$, we have

$$\begin{aligned} sun(G-X)\ge sun(G'-X)-2\ge 2|X|-\varepsilon _2(X)+1-2\ge 2|X|-3=2r-1\ge 1, \end{aligned}$$

which implies that there exists $w\in V(G-X)$ with $d_{G-X}(w)\le 1$, and so

$$\begin{aligned} d_G(w)\le d_{G-X}(w)+|X|\le |X|+1=r+2, \end{aligned}$$

which contradicts that G is $(r+3)$-edge-connected. Hence, $|X|\ge r+2$. This completes the proof of Claim 1. $\square$

Assume that $G'-X$ admits a isolated vertices, b $K_2$’s and c big sun components $H_1,H_2,\cdots ,H_c$, where $|V(H_i)|\ge 6$ for $1\le i\le c$. In terms of (4), $\varepsilon _2(X)\le 2$ and Claim 1, we infer

$$\begin{aligned} sun(G'-X)=a+b+c\ge 2|X|-\varepsilon _2(X)+1\ge 2|X|-1=2r+3. \end{aligned}$$

(5)

Write $R_i$ for the factor-critical graph of $H_i$, $i=1,2,\cdots ,c$. Set $D_i=V(R_i)$ and $D=\bigcup \limits _{i=1}^{c}{D_i}$. Obviously, $i(H_i-D_i)=|D_i|=\frac{|V(H_i)|}{2}\ge 3$ and $|D|\ge 3c$. Select one vertex from each $K_2$ component of $G'-X$, and denote the set of such vertices by Q. We denote by W the union of all non-sun components of $G'-X$.

Note that $i(G'-X)-2\le i(G-X)\le i(G'-X)$. The following proof will be divided into three cases.

Case 1. $i(G-X)=i(G'-X)$.

Clearly, $u,v\notin V(aK_1)$ (otherwise, $i(G-X)<i(G'-X)$, a contradiction).

Subcase 1.1. $u\in V(W)$.

In this subcase, we get $i(G-X-Q-D-u)\ge a+b+|D|$, and so

$$\begin{aligned} I(G)\le \frac{|X\cup Q\cup D\cup \{u\}|}{i(G-X-Q-D-u)}\le \frac{|X|+b+|D|+1}{a+b+|D|}. \end{aligned}$$

Combining this with $I(G)>\frac{3r+6}{2r+3}$, we obtain

$$\begin{aligned} \frac{3r+6}{2r+3}<I(G)\le \frac{|X|+b+|D|+1}{a+b+|D|}, \end{aligned}$$

which implies

$$\begin{aligned} 0>(3r+6)a+(r+3)b+(r+3)|D|-(2r+3)|X|-(2r+3). \end{aligned}$$

(6)

It follows from (5), (6), $|D|\ge 3c$ and Claim 1 that

$$\begin{aligned} 0> & {} (3r+6)a+(r+3)b+(r+3)|D|-(2r+3)|X|-(2r+3)\\\ge & {} (r+3)a+(r+3)b+(r+3)c-(2r+3)|X|-(2r+3)\\= & {} (r+3)(a+b+c)-(2r+3)|X|-(2r+3)\\\ge & {} (r+3)(2|X|-1)-(2r+3)|X|-(2r+3)\\= & {} 3|X|-(3r+6)=3(r+2)-(3r+6)=0, \end{aligned}$$

a contradiction.

Subcase 1.2. $u\notin V(W)$.

In this subcase, $u\in V(bK_2)$, or $u\in D_i$ ($1\le i\le c$), or $u\in V(H_i)\setminus D_i$ ($1\le i\le c$).

Claim 2. $I(G)\le \frac{|X|+b+|D|}{a+b+|D|}$.

Proof

If $u\in V(bK_2)$, then we choose such set Q with $u\in Q$. Thus, we get $i(G-X-Q-D)=a+b+|D|$, and so

$$\begin{aligned} I(G)\le \frac{|X\cup Q\cup D|}{i(G-X-Q-D)}=\frac{|X|+b+|D|}{a+b+|D|}. \end{aligned}$$

If $u\in D_i$ ($1\le i\le c$), then we admit $i(G-X-Q-D)=a+b+|D|$, and so

$$\begin{aligned} I(G)\le \frac{|X\cup Q\cup D|}{i(G-X-Q-D)}=\frac{|X|+b+|D|}{a+b+|D|}. \end{aligned}$$

If $u\in V(H_i)\setminus D_i$ ($1\le i\le c$), then there exists $w\in D_i$ with $uw\in E(H_i)$. Thus, we derive $i(G-X-Q-u-(D\setminus \{w\}))=a+b+|D|$, and so

$$\begin{aligned} I(G)\le \frac{|X\cup Q\cup \{u\}\cup (D\setminus \{w\})|}{i(G-X-Q-u-(D\setminus \{w\}))}=\frac{|X|+b+|D|}{a+b+|D|}. \end{aligned}$$

We finish the proof of Claim 2.

According to Claim 2 and $I(G)>\frac{3r+6}{2r+3}$, we acquire

$$\begin{aligned} \frac{3r+6}{2r+3}<I(G)\le \frac{|X|+b+|D|}{a+b+|D|}, \end{aligned}$$

namely,

$$\begin{aligned} 0>(3r+6)a+(r+3)b+(r+3)|D|-(2r+3)|X|. \end{aligned}$$

(7)

It follows from (5), (7), $|D|\ge 3c$ and Claim 1 that

$$\begin{aligned} 0> & {} (3r+6)a+(r+3)b+(r+3)|D|-(2r+3)|X|\\\ge & {} (r+3)(a+b+c)-(2r+3)|X|\\\ge & {} (r+3)(2|X|-1)-(2r+3)|X|\\= & {} 3|X|-(r+3)=3(r+2)-(r+3)=2r+3>0, \end{aligned}$$

which is a contradiction.

Case 2. $i(G-X)=i(G'-X)-1$.

In this case, $u\in V(aK_1)$ and $v\notin V(aK_1)$, or $u\notin V(aK_1)$ and $v\in V(aK_1)$. Without loss of generality, let $u\in V(aK_1)$ and $v\notin V(aK_1)$. Obviously, $a\ge 1$.

Claim 3. $I(G)\le \frac{|X|+b+|D|+1}{a+b+|D|}$.

Proof

The proof is similar to that of Claim 2 by discussing $v\in V(bK_2)$, $v\in D_i$, $v\in V(H_i)\setminus D_i$ or $v\in V(W)$. Claim 3 is verified. $\square$

According to Claim 3 and $I(G)>\frac{3r+6}{2r+3}$, we have

$$\begin{aligned} \frac{3r+6}{2r+3}<I(G)\le \frac{|X|+b+|D|+1}{a+b+|D|}, \end{aligned}$$

that is,

$$\begin{aligned} (2r+3)|X|>(3r+6)a+(r+3)b+(r+3)|D|-(2r+3). \end{aligned}$$

(8)

In terms of (5), (8), $a\ge 1$ and $|D|\ge 3c$, we deduce

$$\begin{aligned} (2r+3)|X|> & {} (3r+6)a+(r+3)b+(r+3)|D|-(2r+3)\\\ge & {} (3r+6)a+(r+3)b+(r+3)c-(2r+3)\\= & {} (r+3)(a+b+c)+(2r+3)(a-1)\\\ge & {} (r+3)(2|X|-1), \end{aligned}$$

which implies $|X|<\frac{r+3}{3}$, which contradicts Claim 1.

Case 3. $i(G-X)=i(G'-X)-2$.

In this case, $u,v\in V(aK_1)$, and so $a\ge 2$. Thus, we possess $i(G-X-Q-v-D)=a+b+|D|-1$, and so

$$\begin{aligned} I(G)\le \frac{|X\cup Q\cup \{v\}\cup D|}{i(G-X-Q-v-D)}=\frac{|X|+b+|D|+1}{a+b+|D|-1}. \end{aligned}$$

(9)

In light of (9) and $I(G)>\frac{3r+6}{2r+3}$, we derive

$$\begin{aligned} \frac{3r+6}{2r+3}<I(G)\le \frac{|X|+b+|D|+1}{a+b+|D|-1}, \end{aligned}$$

which implies

$$\begin{aligned} (2r+3)|X|>(3r+6)a+(r+3)b+(r+3)|D|-5r-9. \end{aligned}$$

(10)

It follows from (5), (10), $a\ge 2$ and $|D|\ge 3c$ that

$$\begin{aligned} (2r+3)|X|> & {} (3r+6)a+(r+3)b+(r+3)|D|-5r-9\\\ge & {} (3r+6)a+(r+3)b+(r+3)c-5r-9\\= & {} (r+3)(a+b+c)+(2r+3)a-5r-9\\\ge & {} (r+3)(2|X|-1)+2(2r+3)-5r-9\\= & {} 2(r+3)|X|-2r-6, \end{aligned}$$

which implies $|X|<\frac{2r+6}{3}$, which contradicts Claim 1 by $r\ge 1$. Theorem 8 is verified. $\Box$

Remark 3

In what follows, we claim that $I(G)>\frac{3r+6}{2r+3}$ in Theorem 8 cannot be replaced by $I(G)>\frac{3r+5}{2r+3}$.

To explain this, we construct a graph $G=K_{r+3}\vee ((2r+4)K_2)$, where r is a nonnegative integer. We select one vertex from each $K_2$ component of $G-V(K_{r+3})$, and denote the set of such vertices by Y. Then $\frac{3r+6}{2r+3}>I(G)=\frac{|V(K_{r+3})\cup Y|}{i(G-(V(K_{r+3})\cup Y))} =\frac{3r+7}{2r+4}>\frac{3r+5}{2r+3}$ and G is $(r+4)$-edge-connected. Let $G'=G-e$ for $e\in E((2r+4)K_2)$. Let $X=V(K_{r+3})$. Then $|X|=r+3$ and $\varepsilon _2(X)=2$. Thus, we obtain

$$\begin{aligned} sun(G'-X)=2r+5>2r+4=2|X|-\varepsilon _2(X). \end{aligned}$$

In view of Theorem 4, $G'$ is not a $P_{\ge 3}$-factor covered graph, and so G is not a $P_{\ge 3}$-factor uniform graph.

Remark 4

We now claim that 3-edge-connected in Theorem 8 is best possible in some sense.

To show this, we construct a graph $G=K_{r+1}\vee ((2r)K_2)$, where r is an integer with $1\le r\le 2$. We select one vertex from each $K_2$ component of $G-V(K_{r+1})$, and denote the set of such vertices by Y. Then G is 2-edge-connected and $I(G)=\frac{|V(K_{r+1})\cup Y|}{i(G-(V(K_{r+1})\cup Y))}=\frac{3r+1}{2r}>\frac{3r+6}{2r+3}$. Write $G'=G-e$ for $e\in E((2r)K_2)$ and $X=V(K_{r+1})$. Then $|X|=r+1$ and $\varepsilon _2(X)=2$. Hence, we derive

$$\begin{aligned} sun(G'-X)=2r+1>2r=2|X|-\varepsilon _2(X). \end{aligned}$$

By virtue of Theorem 4, $G'$ is not a $P_{\ge 3}$-factor covered graph, and so G is not a $P_{\ge 3}$-factor uniform graph.