All graphs considered here are undirected, connected and simple. Let $ G $ be a graph with the vertex set $ V(G) $ and the edge set $ E(G) $. The order of $ G $ is $ |V(G)| $. For a vertex $ v\in V(G) $, define $ N_G(v)=\{u\in V(G): uv\in E(G)\} $ and $ \Gamma (v)=\{e\in E(G): e \ \text{is incident with} \ v \ \text{in} \ G\} $. An integer $ k $-matching of a graph $ G=(V, E) $ is a function $ h $ that assigns to each edge an integer in $ \{0, \cdots, k\} $ such that $ \sum\limits_{e\in \Gamma (v)} f(e)\leq k $ for each $ v\in V $. A vertex $ v $ of $ G $ is saturated by an integer $ k $-matching $ h $ or $ v $ is $ h $-saturated if $ \sum\limits_{e\in \Gamma (v)} f(e)=k $, otherwise, $ v $ is $ h $-unsaturated. An integer $ k $-matching is perfect if $ \sum\limits_{e\in \Gamma (v)} f(e)=k $ for every vertex $ v\in V(G) $.

Clearly, an integer $ k $-matching is perfect if and only if its size is $ \frac{k|V(G)|}{2} $. Note that when $ k $ is odd, a graph with a perfect integer $ k $-matching has an even number of vertices. For $ k=1 $, the integer (or perfect) 1-matching is just the matching (or perfect matching) in usual sense. If $ k=2 $, then the perfect 2-matching coincides with the one defined by Tutte ([1]).

Integer $ k $-matchings have attracted many researchers' attention, such as Lu et al. ([2]) studied the perfect $ k $-matchings of general graph and gave a sufficient and necessary condition for its existence, which is the Tutte's Theorem for perfect integer $ k $-matching of a graph. Liu et al. ([3]) proved that when $ k $ is even, the integer $ k $-matching number of $ G $ equals $ k $ times its fractional matching number. Hence, by the Berge-Tutte formula of fractional matching in ([4]), they can easily obtained the integer $ k $-matching analogue of the Berge-Tutte Formula when $ k $ is even.

In addition, Gallai (1963, [5]) introduced the concepts of factor-critical graphs, i.e. a graph $ G $ is factor-critical if $ G-v $ has a perfect matching for every vertex $ v $ of $ G $. Very recently, Liu et al. ([6]) extended this definition and defined $ k $-factor-critical graph. A connected graph $ G $ with at least three vertices is said to be $ k $-factor-critical if for any $ v\in V(G) $, there exists an integer $ k $-matching $ h $ such that $ \sum\limits_{e\in \Gamma (v)} f(e)=k-1 $ and other vertices are $ h $-saturated. We note that this definition is different from the one defined by Favaron and Odile ([7]). Moreover, Liu et al. ([6]) proved the integer $ k $-matching analogue of the Berge-Tutte Formula when $ k $ is odd, they also gave sufficient and necessary conditions for the existence of $ k $-factor-critical.

For two subsets $ S, T\subset V(G) $, let $ e(S, T) $ denotes the number of edges of $ G $ joining $ S $ to $ T $. For a set $ X $, we denote the cardinality of $ X $ by $ |X| $. Given a subset $ S\subset V(G) $, we let $ e(S, S^c) $ denote the number of edges with exactly one endpoint in $ S $, where $ S^c $ denotes the complement of $ S $ in $ K_{|V(G)|} $. The expansion constant of a graph $ G $ is $ h(G)=\text{min} \frac{e(S, S^c)}{|S|} $, where the minimum is taken over all $ S\subset V(G) $ with $ |S|\leq \frac{|V(G)|}{2} $, the expansion constant of a graph represents a strength of connection of the graph ([8]).

In [9], Cioab$ \breve{a} $ proved a sufficient condition for the existence of a perfect matching in a regular graph in terms of expansion constant and $ \lambda_3 $, which improved the result of Brouwer and Haemers ([10]). For regular graphs, Lu et al. ([2]) provided a sufficient condition for the existence of perfect $ k $-matching in terms of the edge connectivity. Based on this, we will consider the existence of perfect integer $ k $-matching and $ k $-factor-critical graph in terms of the expansion constant. We present a sufficient condition for the existence of a perfect integer $ k $-matching in a regular graph in terms of its expansion constant, which generalizes the results of Haemers et al. ([10]) and Cioab$ \breve{a} $ et al. ([9]). Furthermore, a sufficient condition for the existence of $ k $-factor-critical graphs for regular graphs in terms of its expansion constant is obtained.

In this section, we determine a lower bound on the expansion constant of a regular graph, which implies the existence of a perfect integer $ k $-matching.

Given a subset $ S\subseteq V(G) $, we say $ G-S $ is the subgraph obtained from G by deleting all vertices of $ S $ and their incident edges. In 2014, Lu et al. ([2]) obtained the Tutte's Theorem for perfect integer $ k $-matching of a graph.

Lemma 2.1 (see [2]) Let $ k \geq 2 $ be even. A graph $ G $ has a perfect integer $ k $-matching if and only if

$ \begin{equation*} i(G-S)\leq |S| \ for \ all \ S\subseteq V(G), \end{equation*} $

where $ i(G-S) $ is the number of isolated vertices of $ G-S $.

Let $ odd(G) $ denote the number of odd components with order at least three in $ G $.

Lemma 2.2 (see [2]) Let $ k \geq 1 $ be odd. A graph $ G $ has a perfect integer $ k $-matching if and only if

$ \begin{equation*} odd(G-S)+k\cdot i(G-S)\leq k|S| \ for \ all \ S\subseteq V(G), \end{equation*} $

where $ odd(G-S) $ is the number of odd components with order at least three of $ G-S $ and $ i(G-S) $ is the number of isolated vertices of $ G-S $.

Theorem 2.1 Let $ G $ be a $ d $-regular graph with $ n $ vertices.

(1) If $ k $ is even, and

$ h(G)\geq \left\{ \begin{array}{ll} \frac{d-2}{d+1}, \ \ \ if \ d \ is \ even, \\ \frac{d-1}{d+1}, \ \ \ otherwise, \end{array} \right. $

then $ G $ has a perfect integer $ k $-matching;

(2) For even $ n $. If $ k $ is odd, and

$ h(G)\geq \left\{ \begin{array}{ll} \frac{d-2}{d+1}, \ \ \ if \ d \ is \ even, \\ \frac{d-2}{d+2}, \ \ \ otherwise, \end{array} \right. $

then $ G $ has a perfect integer $ k $-matching.

Proof (1) Assume $ G $ has no perfect integer $ k $-matching. By Lemma 2.1, there is a vertex set $ S $ such that $ i(G-S) > |S| $. Let $ |S|=s $ and $ i(G-S)=q $. We easily observe $ q\geq s+1 $. Let $ V^{\prime}=V(G)\backslash S $. Thus $ |V^{\prime}|=n-s $. We denote the $ q $ isolated vertices of $ G-S $ by $ v_1, v_2, \cdots, v_q $. Let $ G_1, G_2, \cdots, G_q $ be the subgraphs induced by $ v_i $ and $ N_G(v_i) $, where $ N_G(v_i) $ is the set of neighbours of $ v_i $ and $ 1\leq i\leq q $. Denote by $ n_i $ and $ e_i $ the order and the size of $ G_i $, respectively.

For $ i\in[q] $, let $ t_i $ denote the number of edges in $ G $ between $ G_i $ and $ S-N_G(v_i) $. Since $ G $ is connected, it follows that $ t_i\geq 1 $ for each $ i\in[q] $. Because $ v_i $ is adjacent only to vertices in $ S $, we deduce that $ 2e_i=d(d+1)-t_i $. Clearly $ t_i $ is even.

The sum of the degrees of the vertices in $ S $ is at least the number of edges between $ S $ and $ \cup^q_{i=1}v_i $. Thus, $ ds\geq \sum\limits^q_{i=1} t_i $. Since $ q\geq s+1 $, it follows that there are at least two $ t_i $'s such that $ t_i < d $. This implies there are at least two $ t_i $'s satisfying $ t_i\leq d-2 $ if $ d $ is even and $ t_i\leq d-1 $ if $ d $ is odd.

The fact $ n_1+n_2 < n $ implies that there is at least one $ i $ such that $ n_i < \frac{n}{2} $. Without losing generality, assume $ n_1 < \frac{n}{2} $. Then we obtain $ \frac{t_1}{n_1}\leq \frac{d-2}{d+1} $ if $ d $ is even and $ \frac{t_1}{n_1}\leq \frac{d-1}{d+1} $ if $ d $ is odd. This contradicts the assumption made on the expansion constant, which finishes the proof.

(2) By the definition of perfect integer $ k $-matching, we know that a graph with a perfect integer $ k $-matching has an even number of vertices when $ k $ is odd. So we only discuss the case $ n $ is even.

Assume $ G $ has no perfect integer $ k $-matching. By Lemma 2.2, there is a vertex set $ S $ such that $ odd(G-S)+k\cdot i(G-S) > k|S| $. Let $ |S|=s $ and $ odd(G-S)+i(G-S)=q $. By easily process, we can obtain $ q > s $. But since $ n $ is even, $ s+q $ is even, hence $ q\geq s+2 $. Let $ V^{\prime}=V(G)\backslash S $. Thus $ |V^{\prime}|=n-s $. We denote the $ q $ components by $ G_1, G_2, \cdots, G_q $. Denote by $ n_i $ and $ e_i $ the order and the size of $ G_i $ respectively.

For $ i\in[q] $, let $ t_i $ denotes the number of edges with one endpoint in $ G_i $ and another in $ S $. Since $ G $ is connected, it follows that $ t_i\geq 1 $ for each $ i\in[q] $. Because vertices in $ G_i $ are adjacent only to vertices in $ G_i $ or $ S $, we deduce that $ 2e_i=dn_i-t_i=d(n_i-1)+d-t_i $. Because $ n_i $ is odd, it follows that $ d-t_i $ is even. Hence, $ t_i $ and $ d $ have the same parity for each $ i $.

The sum of the degrees of the vertices in $ S $ is at least the number of edges between $ S $ and $ \cup^q_{i=1}G_i $. Thus, $ ds\geq \sum\limits^q_{i=1} t_i $. Since $ q\geq s+2 $, it follows that there are at least three $ t_i $'s such that $ t_i < d $. This implies there are at least three $ t_i $'s satisfying $ t_i\leq d-2 $. If $ t_i\leq d-2 $, then $ n_i > 1 $. Assume $ t_i\leq d-2 $ for $ i\in[3] $. If $ t_i\leq d-2 $, then $ n_i(n_i-1)\geq 2e_i=dn_i-t_i\geq dn_i-d+2 $. Thus, $ n_i\geq d+\frac{2}{n_i-1} $. Hence, $ n_i\geq d+1 $. Now, since $ d $ and $ n_i $ are both odd, we obtain $ n_i\geq d+2 $.

The fact $ n_1+n_2+n_3 < n $ implies that there is at least one $ i $ such that $ n_i < \frac{n}{2} $. Without losing generality, assume $ n_1 < \frac{n}{2} $. Then we obtain $ \frac{t_1}{n_1}\leq \frac{d-2}{d+1} $ if $ d $ is even and $ \frac{t_1}{n_1}\leq \frac{d-2}{d+2} $ if $ d $ is odd. This contradicts the assumption made on the expansion constant, which finishes the proof.

From the above Theorem 2.1, if $ n $ is even and let $ k=1 $, then we may deduce the following results obtained by Cioab$ \breve{a} $ [9], which provides an expansion constant condition to guarantee that there exists a perfect matching in a regular graph $ G $.

Corollary 2.1 (see [9, 10]) Let $ G $ be a $ d $-regular graph. If $ n $ is even and

$ h(G)\geq \left\{ \begin{array}{ll} \frac{d-2}{d+1}, \ \ \ if \ d \ is \ even, \\ \frac{d-2}{d+2}, \ \ \ if \ d \ is \ odd, \end{array} \right. $

then $ G $ has a perfect matching.

In this section, for a positive integer $ k $ and a regular graph $ G $, we obtain a condition of $ k $-factor-critical according to a lower bound of the expansion constant $ h(G) $.

In 2021, Liu et al. ([6]) gave a sufficient and necessary condition for the existence of $ k $-factor-critical graph.

Lemma 2.3 (see [6]) A connected graph $ G $ with at least three vertices is $ k $-factor-critical if and only if $ G $ has odd number of vertices and $ odd(G-S)+k\cdot i(G-S)\leq k|S|-1 $ for any $ \emptyset \neq S\subseteq V(G) $.

Theorem 2.2 Let $ G $ be a $ d $-regular graph with $ n $ vertices. If $ n $ is odd and

$ h(G)\geq \left\{ \begin{array}{ll} \frac{d-2}{d+1}, \ \ \ if \ d \ is \ even, \\ \frac{d-2}{d+2}, \ \ \ if \ d \ is \ odd, \end{array} \right. $

then $ G $ is $ k $-factor-critical.

Proof Suppose $ G $ satisfies the conditions of this theorem, but it is not $ k $-factor-critical. By Lemma 2.3, there is a vertex set $ S $ such that $ odd(G-S)+k\cdot i(G-S)\geq k|S| $. Let $ |S|=s $ and $ odd(G-S)+i(G-S)=q $, we can get $ q\geq s $. And since $ n $ is odd, $ s+q $ is odd, hence $ q\geq s+1 $.

Let $ G_1, G_2, \cdots, G_q $ be the $ q $ odd components with odd vertices of $ G-S $. We denote the order and size of $ G_{i} $ by $ n_{i} $ and $ e_{i} $, respectively. For $ i\in[q] $, let $ t_i $ denote the number of edges with one endpoint in $ G_i $ and another in $ S $. We know $ t_i\geq 1 $ for $ i\in[q] $ because $ G $ is connected. Since vertices in $ G_{i} $ are adjacent only to vertices in $ G_{i} $ or $ S $, we can get that $ 2e_i=dn_i-t_i=d(n_i-1)+d-t_i $. This implies that $ d-t_{i} $ is even since $ n_{i} $ is odd. Therefore, $ t_{i} $ and $ d $ have the same parity for each $ i $.

The sum of degrees of the vertices of $ S $ is at least the number of edges between $ S $ and the odd component $ G_{i} $ for $ i\in [q] $. Hence, $ ds\geq \sum\limits^q_{i=1} t_i $. Because $ q\geq s+1 $, it follows that there are at least two $ t_i $'s such that $ t_i < d $. Hence, there are at least two $ t_i $'s such that $ t_i < d-2 $. If $ t_{i}\leq d-2 $, then $ n_{i} > 1 $. Without loss of generality, we assume $ t_{i}\leq d-2 $ for $ i\in [2] $, then $ n_{i}(n_{i}-1)\geq 2e_{i}=dn_{i}-t_{i} $. Hence, $ n_{i}\geq d+\frac{2}{n_{i}-1} $. Then $ n_{i}\geq d+1 $. If $ d $ is odd, we get $ n_{i}\geq d+2 $ since $ n_{i} $ is odd.

There is at least one $ i $ such that $ n_{i} < \frac{n}{2} $ because of $ n_{1}+n_{2} < n $. We assume $ n_{1} < \frac{n}{2} $. Therefore, we get $ \frac{t_{1}}{n_{1}}\leq \frac{d-2}{d+1} $ if $ d $ is even and $ \frac{t_{1}}{n_{1}}\leq \frac{d-2}{d+2} $ if $ d $ is odd. This contradicts the condition made on the expansion constant, which completes the proof.