All graphs considered are simple, finite and undirected. Let G be a graph. The degree of a vertex v of G is denoted by $d_G\left(v\right)$. The neighbour set of a vertex subset S of G is the set of vertices adjacent to a vertex in S, denoted by $N_G\left(S\right)$. For two subsets E₁ and E₂ of $E\left(G\right)$, the symmetric difference of E₁ and E₂ is denoted by $E_1\Delta E_2$, that is, $E_1\Delta E_2=\left(E_1-E_2\right)\cup \left(E_2-E_1\right)$. For two subsets $S,T$ of $V\left(G\right)$, let $E_G\left(S,T\right)=\left\{uv\in E\left(G\right):u\in S,v\in T\right\}$. The complete bipartite graph with bipartition $\left(X,Y\right)$ with that $\left|X\right|=s$ and $\left|Y\right|=t$ is denoted by $K_{s,t}$. Specially, we denote by $K_{1,0}$ a single vertex. We refer to the book [1] for graph theoretical notations and terminology that are not defined here.

Let $G=\left(X,Y\right)$ be a bipartite graph with bipartition $\left(X,Y\right)$. A semi-matching of G is defined as a set of edges $M\subseteq E\left(G\right)$ such that each vertex in Y is incident with exactly one edge in M, and a vertex in X can be incident with an arbitrary number of edges in M. In general, valid semi-matchings of a connected graph G can be easily obtained by matching each vertex $y\in Y$ with an arbitrary vertex $x\in X$ for which $xy\in E\left(G\right)$. In the computer science, the problem about the semi-matchings in a bipartite graph $G=\left(X,Y\right)$ is well known (see [2] ), where Y corresponds to the clients (or tasks), X to the servers (or machines) and the number of edges in a semi-matching incident with a server (machine) in X is seen as the load on the server (machine). So semi-matchings are widely considered on different optimization objectives (see [3] - [6] ). Many optimization objectives are to find the fairest semi-matching related to the load-balancing problem in a system, where a set of tasks need to be assigned to a set of machines in the fairest way. An example of the fairest semi-matching according to Jain’s index is shown in Figure 1 [5] and M₄ is the fairest one with the highest index of 0.86. Clearly, if $G=\left(X,Y\right)$ has a semi-matching M such that every vertex in X has the same load k, then M is a fairest semi-matching of G, which is said to be a perfect 1-k matching. Therefore, It is significant to study the new problem about 1-k matchings of a bipartite graph.

Now we give the definition of 1-k matchings. Let k be a positive integer. A 1-k matching with respect to X is an edge subset M of G such that each vertex in Y is incident with at most one edge in M and each vertex in X either is incident with exactly k edges in M or is not incident with any edge in M. In the following, 1-k matchings refer to ones with respect to X. A vertex is called M-saturated if it is incident with an edge in M, otherwise, it is M-unsaturated. A 1-k matching M is perfect if every vertex is M-saturated, that is, M covers every vertex of G. Then a perfect 1-k matching is an ideal state of semi-matchings which optimize some objectives.

When $k=1$, a 1-k matching is a matching. When $k=2$, Izumi and Watanabe [7] studied maximum 1 - 2 matching by giving augmenting trail, where augmenting trail is a generalization of the augmenting path for matchings and presented an algorithm for finding a maximum 1 - 2 matching in bipartite graphs.

Our work is to extend the matching theory on bipartite graphs to 1-k matching, motivated by the classical matching problem. Hall’s theorem is well known for judging the existence of perfect matchings [8] (see Theorem 1.1) and the deficient form of Hall’s theorem is given in [9] (see Corollary 1.1). Moreover, in the matching theory [8] , the elementary bipartite graphs about matching are well characterized (see Proposition 1.1). An edge of a graph G is allowed if it lies in some perfect matching of G, otherwise, it is forbidden. A graph G is said to be elementary if its allowed edges form a connected spanning subgraph of G, that is, the subgraph obtained from G by deleting all forbidden edges is connected.

Theorem 1.1. (Hall’s Theorem [8] ) A bipartite graph $G=\left(X,Y\right)$ has a matching which covers every vertex in X if and only if

$\left|N_G\left(S\right)\right|\ge \left|S\right|$

for any $S\subseteq X$.

Corollary 1.1. [9] A bipartite graph $G=\left(X,Y\right)$ has a matching which covers $\left|X\right|-d$ vertices in X if and only if

$\left|N_G\left(S\right)\right|\ge \left|S\right|-d$

for any $S\subseteq X$.

Proposition 1.1. [8] Let G be a bipartite graph with bipartition $\left(X,Y\right)$. Then the following statements are equivalent:

1) G is elementary;

2) G has exactly two minimum vertex covers, named X and Y;

3) $\left|X\right|=\left|Y\right|$ and for every non-empty proper subset S of X, $\left|N_G\left(S\right)\right|\ge \left|S\right|+1$;

4) $G=K_2$, or $\left|V\left(G\right)\right|\ge 4$ and for any $u\in X$ and $v\in Y$, $G-u-v$ has a perfect matching;

5) G is connected and every edge of G is allowed.

In this paper, we give three sufficient and necessary conditions for the existence of perfect 1-k matchings and for the existence of 1-k matchings covering $\left|X\right|-d$ vertices in X, respectively. Moreover, we characterize k-elementary bipartite graph, which is a connected graph such that the subgraph induced by all k-allowed edges is connected.

We give the following theorem about 1-k matching similar to Hall’s theorem.

Theorem 2.1. A bipartite graph $G=\left(X,Y\right)$ has a 1-k matching, which covers every vertex in X if and only if

$\left|N_G\left(S\right)\right|\ge k\left|S\right|$ for any $S\subseteq X$.(1)

Proof. It is obvious that if there exists a 1-k matching covering X, then the condition (1) holds. Now we prove “sufficiency” by induction on $\left|X\right|$. Suppose that $\left|N_G\left(S\right)\right|\ge k\left|S\right|$ for any $S\subseteq X$. When $\left|X\right|=1$, we have that $\left|N_G\left(X\right)\right|\ge k$. Then G has a 1-k matching covering X. Suppose that $\left|X\right|\ge 2$. We consider the following three cases.

Case 1 There exists a non-empty proper subset Z of X such that $\left|N_G\left(Z\right)\right|=k\left|Z\right|$.

Let H₁ be the subgraph induced by $Z\cup N_G\left(Z\right)$. Then H₁ satisfies the condition (1). By the induction hypothesis, H₁ contains a 1-k matching, say M₁, which covers every vertex in Z. Let H₂ be a subgraph induced by $\left(X-Z\right)\cup \left(Y-N_G\left(Z\right)\right)$.

Claim 1 H₂ satisfies the condition (1).

Suppose, to the contrary, that there exists $S\subseteq X-Z$ such that $\left|N_{H_2}\left(S\right)\right|<k\left|S\right|$. Then

$\left|N_G\left(S\cup Z\right)\right|=\left|N_G\left(Z\right)\right|+\left|N_{H_2}\left(S\right)\right|<k\left|Z\right|+k\left|S\right|=k\left|Z\cup S\right|$,

which contradicts with the condition (1).

By Claim 1 and the induction hypothesis, there exists a 1-k matching of H₂, say M₂, which covers every vertex in $X-Z$. Then $M_1\cup M_2$ is a 1-k matching of G covering every vertex in X.

Case 2 For any non-empty proper subset S of X, $\left|N_G\left(S\right)\right|\ge k\left|S\right|+1$.

We choose a non-empty proper subset S₀ of X such that $\left|N_G\left(S_0\right)\right|-k\left|S_0\right|$ is smallest. Let $j=\left|N_G\left(S_0\right)\right|-k\left|S_0\right|$. Then $j\ge 1$ and for any non-empty proper subset S of X, we have that $\left|N_G\left(S\right)\right|\ge k\left|S\right|+j$. It is clear that $N_G\left(S_0\right)\subseteq N_G\left(X\right)$ and $N_G\left(X\right)-N_G\left(S_0\right)\subseteq N_G\left(X-S_0\right)$. By the condition (1), $\left|N_G\left(X\right)\right|\ge k\left|X\right|$. Then

$\left|N_G\left(X\right)-N_G\left(S_0\right)\right|=\left|N_G\left(X\right)\right|-\left|N_G\left(S_0\right)\right|\ge k\left|X-S_0\right|-j$(2)

Subcase 1 $\left|N_G\left(X\right)-N_G\left(S_0\right)\right|\ge k\left|X-S_0\right|$.

Let H₁ be the subgraph induced by $S_0\cup N_G\left(S_0\right)$ and H₂ the subgraph induced by $\left(X-S_0\right)\cup \left(N_G\left(X\right)-N_G\left(S_0\right)\right)$. It is clear that H₁ satisfies the condition (1). By the induction hypothesis, H₁ has a 1-k matching M₁ covering every vertex in S₀.

Claim 2 H₂ satisfies the condition (1).

Clearly, $N_{H_2}\left(X-S_0\right)\subseteq N_G\left(X\right)-N_G\left(S_0\right)$. Let $y\in N_G\left(X\right)-N_G\left(S_0\right)$. Then there exists a vertex $x\in X-S_0$ such that $xy\in E\left(G\right)$. Then $xy\in E\left(H_2\right)$. Hence $y\in N_{H_2}\left(X-S_0\right)$. Thus $N_G\left(X\right)-N_G\left(S_0\right)=N_{H_2}\left(X-S_0\right)$. So we have that

$\left|N_{H_2}\left(X-S_0\right)\right|=\left|N_G\left(X\right)-N_G\left(S_0\right)\right|\ge k\left|X-S_0\right|.$

Suppose, to the contrary, that there exists a non-empty proper subset S₁ of $X-S_0$ such that $\left|N_{H_2}\left(S_1\right)\right|<k\left|S_1\right|$. Let $S=S_0\cup S_1$. Then

$\left|N_G\left(S\right)\right|=\left|N_G\left(S_0\right)\right|+\left|N_{H_2}\left(S_1\right)\right|<k\left|S_0\right|+j+k\left|S_1\right|=k\left|S\right|+j,$

which contradicts with that j is smallest.

By Claim 2 and the induction hypothesis, H₂ has a 1-k matching M₂ covering every vertex in $X-S_0$. Hence $M_1\cup M_2$ is a 1-k matching of G covering every vertex in X.

Subcase 2 $\left|N_G\left(X\right)-N_G\left(S_0\right)\right|<k\left|X-S_0\right|$.

Let $T_1=N_G\left(X-S_0\right)\cap N_G\left(S_0\right)$ and $T_2=N_G\left(X-S_0\right)-T_1$. Then $T_2=N_G\left(X\right)-N_G\left(S_0\right)$. According to (2) and the condition in the case, we have that

$k\left|X-S_0\right|-j\le \left|T_2\right|<k\left|X-S_0\right|.$

Thus $1\le k\left|X-S_0\right|-\left|T_2\right|\le j$. Since

$\left|T_1\right|+\left|T_2\right|=\left|N_G\left(X-S_0\right)\right|\ge k\left|X-S_0\right|+j>\left|T_2\right|+j,$

we have that $\left|T_1\right|\ge j+1$. Then we can choose a set $Z\subseteq T_1$ such that $\left|Z\right|=k\left|X-S_0\right|-\left|T_2\right|$. Hence $1\le \left|Z\right|\le j$. Let H₁ be the subgraph induced by $S_0\cup \left(N_G\left(S_0\right)-Z\right)$ and H₂ the subgraph induced by $\left(X-S_0\right)\cup \left(T_2\cup Z\right)$.

Claim 3 H₁ and H₂ satisfy the condition (1).

Let $\varnothing \ne S^{\prime }\subseteq S_0$. Then

$\left|N_{H_1}\left(S^{\prime }\right)\right|\ge \left|N_G\left(S^{\prime }\right)\right|-\left|Z\right|\ge k\left|S^{\prime }\right|+j-\left|Z\right|\ge k\left|S^{\prime }\right|.$

Hence H₁ satisfies the condition (1). Now we prove that H₂ also satisfies the condition (1). Clearly, $\left|N_{H_2}\left(X-S_0\right)\right|=\left|N_G\left(X\right)-N_G\left(S_0\right)\right|+\left|Z\right|=\left|T_2\right|+\left|Z\right|=k\left|X-S_0\right|$. Suppose, to the contrary, that there exists a non-empty proper subset $S_1$ of $X-S_0$ such that $\left|N_{H_2}\left(S_1\right)\right|<k\left|S_1\right|$. Let $S=S_0\cup S_1$. Then

$\left|N_G\left(S\right)\right|\le \left|N_G\left(S_0\right)\right|+\left|N_{H_2}\left(S_1\right)\right|<k\left|S_0\right|+j+k\left|S_1\right|=k\left|S\right|+j,$

which contradicts with that j is smallest.

Hence by Claim 3 and the induction hypothesis, H₁ has a 1-k matching M₁ covering S₀ and H₂ has a 1-k matching M₂ covering every vertex in $X-S_0$. Hence $M_1\cup M_2$ is a 1-k matching of G covering every vertex in X.

A bipartite graph $G=\left(X,Y\right)$ is called k-balanced with respect to X if $\left|Y\right|=k\left|X\right|$. Now, we give the sufficiency and necessary conditions for the existence of perfect 1-k matchings.

Theorem 2.2. Let $G=\left(X,Y\right)$ be a k-balanced (with respect to X) bipartite graph. Then the following statements are equivalent:

1) G has a perfect 1-k matching;

2) $\left|N_G\left(S\right)\right|\ge k\left|S\right|$ for any $S\subseteq X$;

3) $\left|N_G\left(T\right)\right|\ge \frac{1}{k}\left|T\right|$ for any $T\subseteq Y$;

4) $\left|N_G\left(T\right)\right|\ge \frac{1}{k}\left|T\right|$ for any $T\subseteq Y$ such that $\left|T\right|\equiv 1\left(\mathrm{mod}k\right)$.

Proof. “(1) $\iff$ (2)”. By Theorem 2.1, it is obvious.

“(2) $\Rightarrow$ (3)”. Let T be a subset of Y. If $T=\varnothing$, then $\left|N_G\left(T\right)\right|=0=\frac{1}{k}\left|T\right|$. Suppose that $T\ne \varnothing$. Let $S=X-N_G\left(T\right)$. Then

$k\left|X\right|-k\left|N_G\left(T\right)\right|=k\left|S\right|\le \left|N_G\left(S\right)\right|\le \left|Y\right|-\left|T\right|.$

Since $\left|Y\right|=k\left|X\right|$, we have that $\left|N_G\left(T\right)\right|\ge \frac{1}{k}\left|T\right|$.

“(3) $\Rightarrow$ (2)”. Let S be a subset of X. If $S=\varnothing$, then $\left|N_G\left(S\right)\right|=0=k\left|S\right|$. Suppose that $S\ne \varnothing$. Let $T=Y-N_G\left(S\right)$. Then

$\frac{1}{k}\left|Y\right|-\frac{1}{k}\left|N_G\left(S\right)\right|=\frac{1}{k}\left|T\right|\le \left|N_G\left(T\right)\right|\le \left|X\right|-\left|S\right|.$

Since $\left|Y\right|=k\left|X\right|$, we have that $\left|N_G\left(S\right)\right|\ge k\left|S\right|$.

“(3) $\Rightarrow$ (4)”. It is obvious.

“(4) $\Rightarrow$ (3)”. Let T be a subset of Y. If $T=\varnothing$, then $\left|N_G\left(T\right)\right|=0=\frac{1}{k}\left|T\right|$.

Suppose $T\ne \varnothing$. Let $\left|T\right|=km+c$, where $m\ge 0$ and $0\le c\le k-1$. We distinguish the following cases.

Case 1 $c=0$.

Then $m\ge 1$. Hence we can choose a subset $Z\subseteq T$ such that $\left|Z\right|=k\left(m-1\right)+1$. Then $\left|Z\right|\equiv 1\left(\mathrm{mod}k\right)$. Hence

$\left|N_G\left(T\right)\right|\ge \left|N_G\left(Z\right)\right|\ge \frac{1}{k}\left|Z\right|=\frac{1}{k}\left(km-k+1\right).$

Thus $\left|N_G\left(T\right)\right|\ge m=\frac{1}{k}\left|T\right|$.

Case 2 $c\ne 0$.

Without loss of generality, suppose that $2\le c\le k-1$. Then we can choose a set $Z\subseteq T$ such that $\left|Z\right|=km+1$. Hence

$\left|N_G\left(T\right)\right|\ge \left|N_G\left(Z\right)\right|\ge \frac{1}{k}\left|Z\right|=\frac{1}{k}\left(km+1\right).$

Thus $\left|N_G\left(T\right)\right|\ge m+1>\frac{1}{k}\left|T\right|$.

In the following, we consider the deficient form of Theorem 2.2, similar to the deficient form of Hall’s theorem. If every component of a graph is $K_{s,t}$ and the number of components is m, then the graph is denoted by $m\cdot K_{s,t}$. Let $G=\left(X,Y\right)$ be a bipartite graph and $\mathscr{H}=\left\{K_{1,j}:0\le j\le k\right\}$. An $\mathscr{H}$-quasi factor with respect to X of G is a subgraph H of G such that $V\left(H\right)\cap X=X$ and every component of H is isomorphic to a member $K_{1,j}$ in $\mathscr{H}$ such that the vertex

with degree j in $K_{1,j}$ is in X. Then we can assume that $H={\displaystyle \underset{0\le j\le k}{\cup }{t}_j\cdot K_{1,j}}$, where $t_j\ge 0$. Thus $\underset{j=0}{\overset{k}{{\displaystyle \sum }}}{t}_j=\left|X\right|$.

Corollary 2.1. Let $G=\left(X,Y\right)$ be a k-balanced (with respect to X) bipartite graph. Then the following statements are equivalent:

1) G has an $\mathscr{H}$-quasi factor $H={\displaystyle \underset{0\le j\le k}{\cup }{t}_j\cdot K_{1,j}}$ with respect to X such that $t_k\ge \left|X\right|-d$ and $\underset{j=0}{\overset{k}{{\displaystyle \sum }}}j{t}_j\ge \left|Y\right|-d$, which implies that G has a 1-k matching covering at least $\left|X\right|-d$ vertices in X;

2) $\left|N_G\left(S\right)\right|\ge k\left|S\right|-d$ for any $S\subseteq X$;

3) $\left|N_G\left(T\right)\right|\ge \frac{1}{k}\left(\left|T\right|-d\right)$ for any $T\subseteq Y$;

4) $\left|N_G\left(T\right)\right|\ge \frac{1}{k}\left(\left|T\right|-d\right)$ for any $T\subseteq Y$ such that $\left|T\right|\equiv d+1\left(\mathrm{mod}k\right)$.

Proof. “(1) $\Rightarrow$ (2)”. According to the definition of H, $\underset{j=0}{\overset{k}{{\displaystyle \sum }}}{t}_j=\left|X\right|$. For any $S\subseteq X$, let $S=S_0\cup S_1\cup \cdots \cup S_k$, where $S_j\subseteq V\left[t_j\cdot K_{1,j}\right]$. Then $\left|S_j\right|\le t_j$. Hence we have that

$\begin{array}{c}\left|N_G\left(S\right)\right|\ge \left|N_H\left(S\right)\right|=\underset{i=0}{\overset{k}{{\displaystyle \sum }}}j\left|S_j\right|=\underset{i=0}{\overset{k}{{\displaystyle \sum }}}\left(k-\left(k-j\right)\right)\left|S_j\right|\\ =k\left|S\right|-\underset{i=0}{\overset{k}{{\displaystyle \sum }}}\left(k-j\right)\left|S_j\right|\ge k\left|S\right|-\underset{i=0}{\overset{k}{{\displaystyle \sum }}}\left(k-j\right)t_j\\ =k\left|S\right|-k\underset{j=0}{\overset{k}{{\displaystyle \sum }}}{t}_j+\underset{j=0}{\overset{k}{{\displaystyle \sum }}}j{t}_j\\ \ge k\left|S\right|-k\left|X\right|+\left|Y\right|-d=k\left|S\right|-d.\end{array}$

“(2) $\Rightarrow$ (1)”. Adding d new vertices $y_1,\cdots ,y_d$ to Y of G and joining them to each vertex in X, the resulting graph is denoted by G₁. It is easy to check that G₁ satisfies the condition (1) of the Theorem 2.1. Hence G₁ has a 1-k matching, say M, which covers every vertex in X. Let G₂ be the subgraph induced by M. Then $G_2=\left|X\right|\cdot K_{1,k}$. Let $H=G_2-\left\{y_1,\cdots ,y_d\right\}$. Then H is an $\mathscr{H}$-quasi factor with respect to X of G, H has at least $\left|X\right|-d$ components which are $K_{1,k}$,

$V\left(H\right)\cap X=X$ and $\left|V\left(H\right)\cap Y\right|\ge k\left|X\right|-d=\left|Y\right|-d$. Hence we can assume that

$H={\displaystyle \underset{j=0}{\overset{k}{\cup }}{t}_j\cdot K_{1,j}}$ and $\underset{j=0}{\overset{k}{{\displaystyle \sum }}}j{t}_j=\left|V\left(H\right)\cap Y\right|\ge \left|Y\right|-d$ and $t_k\ge \left|X\right|-d$.

“(2) $\Rightarrow$ (3)”. Let T be a subset of Y. If $T=\varnothing$, then $\left|N_G\left(T\right)\right|\ge 0\ge \frac{1}{k}\left(\left|T\right|-d\right)$. Suppose that $T\ne \varnothing$. Let $S=X-N_G\left(T\right)$. Then

$k\left|X\right|-k\left|N_G\left(T\right)\right|-d=k\left|S\right|-d\le \left|N_G\left(S\right)\right|\le \left|Y\right|-\left|T\right|.$

Since $\left|Y\right|=k\left|X\right|$, we have that $\left|N_G\left(T\right)\right|\ge \frac{1}{k}\left(\left|T\right|-d\right)$.

“(3) $\Rightarrow$ (2)”. Let S be a subset of X. If $S=\varnothing$, then $\left|N_G\left(S\right)\right|\ge 0\ge k\left|S\right|$. Suppose that $S\ne \varnothing$. Let $T=Y-N_G\left(S\right)$. Then

$\frac{1}{k}\left(\left|Y\right|-\left|N_G\left(S\right)\right|-d\right)=\frac{1}{k}\left(\left|T\right|-d\right)\le \left|N_G\left(T\right)\right|\le \left|X\right|-\left|S\right|.$

Since $\left|Y\right|=k\left|X\right|$, we have that $\left|N_G\left(S\right)\right|\ge k\left|S\right|-d$.

“(3) $\Rightarrow$ (4)”. It is obvious.

“(4) $\Rightarrow$ (3)”. Let T be a subset of Y. If $\left|T\right|\le d$, then $\left|N_G\left(T\right)\right|\ge 0\ge \frac{1}{k}\left(\left|T\right|-d\right)$. Suppose that $\left|T\right|>d$. Let $\left|T\right|-d=km+c$, where $m\ge 0$ and $0\le c\le k-1$. We distinguish the following cases.

Case 1 $c=0$.

Then $m\ge 1$. Hence we can choose a subset $Z\subseteq T$ such that $\left|Z\right|-d=k\left(m-1\right)+1$. Hence

$\left|N_G\left(T\right)\right|\ge \left|N_G\left(Z\right)\right|\ge \frac{1}{k}\left(\left|Z\right|-d\right)=\frac{1}{k}\left(km-k+1\right).$

Thus $\left|N_G\left(T\right)\right|\ge m=\frac{1}{k}\left(\left|T\right|-d\right)$.

Case 2 $c\ne 0$.

Without loss of generality, suppose that $2\le c\le k-1$. Then we can choose a set $Z\subseteq T$ such that $\left|Z\right|-d=km+1$. Hence

$\left|N_G\left(T\right)\right|\ge \left|N_G\left(Z\right)\right|\ge \frac{1}{k}\left(\left|Z\right|-d\right)=\frac{1}{k}\left(km+1\right).$

Thus $\left|N_G\left(T\right)\right|\ge m+1>\frac{1}{k}\left(km+c\right)=\frac{1}{k}\left(\left|T\right|-d\right)$.

A edge of a graph G is k-allowed if it lies in some perfect 1-k matching of G, otherwise, it is k-forbidden. A bipartite graph is said to be a k-elementary graph if its k-allowed edges form a connected spanning subgraph of G. Now we characterize k-elementary bipartite graphs.

Theorem 3.1. Let $G=\left(X,Y\right)$ be a bipartite graph. Then the following statements are equivalent:

1) G is k-elementary;

2) G is connected, k-balanced (with respect to X) and $\left|N_G\left(S\right)\right|>k\left|S\right|$ for every non-empty proper subset S of X;

3) G is connected, k-balanced (with respect to X) and $\left|N_G\left(T\right)\right|>\frac{1}{k}\left|T\right|$ for every non-empty proper subset T of Y;

4) G is connected, k-balanced (with respect to X) and $\left|N_G\left(T\right)\right|>\frac{1}{k}\left|T\right|$ for every non-empty proper subset T of Y such that $\left|T\right|\equiv 0,1\left(\mathrm{mod}k\right)$;

5) G is connected and every edge of G is k-allowed;

6) $G=K_{1,k}$, or $\left|V\left(G\right)\right|\ge 2k+2$ and for any $x\in X$ and $y\in Y$, there exist $y_1,y_2,\cdots ,y_{k-1}\in N_G\left(x\right)-\left\{y\right\}$ such that $G-\left\{x,y,y_1,y_2,\cdots ,y_{k-1}\right\}$ has a perfect 1-k matching.

Proof. “(1) $\Rightarrow$ (2)”. Since G is k-elementary, then G is connected and has a perfect 1-k matching. Then $\left|Y\right|=k\left|X\right|$ and by Theorem 2.1, $\left|N_G\left(S\right)\right|\ge k\left|S\right|$ for any $S\subseteq X$. Then G is k-balanced with respect to X. Suppose, to the contrary, that there exists a non-empty proper subset S₀ of X such that $\left|N\left(S_0\right)\right|=k\left|S_0\right|$. Since G is k-elementary, there exists a k-allowed edge of G, say uv, such that $u\in X-S_0$ and $v\in N_G\left(S_0\right)$. Hence we can assume that there exists a perfect 1-k matching M of G containing uv. Let H be the subgraph of G induced by $S_0\cup \left(N_G\left(S_0\right)-\left\{v\right\}\right)$. Then $M\cap E\left(H\right)$ is a 1-k matching of H covering every vertex in S₀. But $\left|N_H\left(S_0\right)\right|=k\left|S_0\right|-1$, which contradicts with Theorem 2.1.

“(2) $\Rightarrow$ (3)”. Let T be a non-empty proper subset of Y and $S=X-N_G\left(T\right)$. If $N_G\left(T\right)=X$, then $\left|N_G\left(T\right)\right|=\left|X\right|=\frac{1}{k}\left|Y\right|>\frac{1}{k}\left|T\right|$. Suppose that $N_G\left(T\right)\subset X$. Since G is connected, $N_G\left(T\right)\ne \varnothing$. Then $S\ne \varnothing$ and $S\subset X$. Hence

$k\left|X\right|-k\left|N_G\left(T\right)\right|=k\left|S\right|<\left|N_G\left(S\right)\right|\le \left|Y\right|-\left|T\right|.$

Since $\left|Y\right|=k\left|X\right|$, we have that $\left|N_G\left(T\right)\right|>\frac{1}{k}\left|T\right|$.

“(3) $\Rightarrow$ (2)”. Let S be a non-empty proper subset of X and $T=Y-N_G\left(S\right)$. If $N_G\left(S\right)=Y$, then $\left|N_G\left(S\right)\right|=\left|Y\right|=k\left|X\right|>k\left|S\right|$. Suppose that $N_G\left(S\right)\subset Y$. Since G is connected, $N_G\left(S\right)\ne \varnothing$. Then $T\ne \varnothing$ and $T\subset Y$. Hence

$\frac{1}{k}\left|Y\right|-\frac{1}{k}\left|N_G\left(S\right)\right|=\frac{1}{k}\left|T\right|<\left|N_G\left(T\right)\right|\le \left|X\right|-\left|S\right|.$

Since $\left|Y\right|=k\left|X\right|$, we have that $\left|N_G\left(S\right)\right|>k\left|S\right|$.

“(3) $\Rightarrow$ (4)”. It is obvious.

“(4) $\Rightarrow$ (3)”. Let T be a non-empty proper subset of Y and $\left|T\right|=km+c$, where $m\ge 0$ and $0\le c\le k-1$. Without loss of generality, suppose that $2\le c\le k-1$. Then we can choose a set $Z\subseteq T$ such that $\left|Z\right|=km+1$. Hence

$\left|N_G\left(T\right)\right|\ge \left|N_G\left(Z\right)\right|>\frac{1}{k}\left|Z\right|=\frac{1}{k}\left(km+1\right).$

Thus $\left|N_G\left(T\right)\right|\ge m+1>\frac{1}{k}\left|T\right|$.

“(2) $\Rightarrow$ (5)”. We prove that every edge is k-allowed. Let $v\in Y$, $uv\in E\left(G\right)$ and $G_0=G-E_G\left(\left\{v\right\},X-\left\{u\right\}\right)$. Then $d_{G_0}\left(v\right)=1$. According to the condition (2), $d_G\left(x\right)\ge k+1$ for any $x\in X$. Since G is connected, $N_{G_0}\left(X\right)=N_G\left(X\right)=Y$. Since G is k-balanced, $\left|N_{G_0}\left(X\right)\right|=\left|Y\right|=k\left|X\right|$. Suppose, to the contrary, that there exists a non-empty proper subset S of X such that $\left|N_{G_0}\left(S\right)\right|\le k\left|S\right|-1$. Then

$\left|N_G\left(S\right)\right|\le \left|N_{G_0}\left(S\right)\right|+1\le k\left|S\right|-1+1=k\left|S\right|,$

a contradiction. By Theorem 2.2, there exists a perfect 1-k matching M of G₀. Then $uv\in M$. Clearly, M is a perfect 1-k matching M of G. So every edge of G is k-allowed.

“(5) $\Rightarrow$ (1)”. It is obvious.

“(6) $\Rightarrow$ (2)”. It implies that G is k-balanced with respect to X and has a perfect 1-k matching. First, we prove that G is connected. Suppose, to the contrary, that G is disconnected. Let $G_1=\left(X_1,Y_2\right),\cdots ,G_s=\left(X_s,Y_s\right)$ be all components of G. Then every $G_i$ has a perfect 1-k matching. Hence $\left|Y_i\right|=k\left|X_i\right|$ for any $1\le i\le s$. Let $x\in X_1$ and $y\in Y_2$. It is clear that for any $y_1,\cdots ,y_{k-1}\in N_G\left(x\right)-\left\{y\right\}$, $G-\left\{x,y,y_1,\cdots ,y_{k-1}\right\}$ has no perfect 1-k matchings, a contradiction. So G is connected. Secondly, we prove that $\left|N_G\left(S\right)\right|>k\left|S\right|$ for every non-empty proper subset S of X. By Theorem 2.2, $\left|N_G\left(S\right)\right|\ge k\left|S\right|$. Suppose, to the contrary, that there exists a non-empty proper subset S₀ of X such that $\left|N_G\left(S_0\right)\right|=k\left|S_0\right|$. Let $x\in X-S_0$ and $y\in N_G\left(S_0\right)$. Then there exist $y_1,\cdots ,y_{k-1}\in N_G\left(x\right)-\left\{y\right\}$ such that $G-\left\{x,y,y_1,\cdots ,y_{k-1}\right\}$ has a perfect 1-k matching. Let $H=G-\left\{x,y,y_1,\cdots ,y_{k-1}\right\}$. Then

$\left|N_H\left(S_0\right)\right|\le k\left|S_0\right|-1,$

which contradicts with Theorem 2.2.

“(3) $\Rightarrow$ (6)”. Since G is connected and k-balanced with respect to X, we have that $\left|N_G\left(Y\right)\right|=\left|X\right|=\frac{1}{k}\left|Y\right|$. By Theorem 2.2, G has a perfect 1-k matching, say M.

When $\left|X\right|=1$, $G=K_{1,k}$. Suppose that $\left|X\right|\ge 2$. Let H be the subgraph induced by M. Let $x\in X$ and $y\in Y$.

Case 1 $xy\in E\left(H\right)$

Let $y_1,\cdots ,y_{k-1}\in N_H\left(x\right)-\left\{y\right\}$. Then $H-\left\{x,y,y_1,\cdots ,y_{k-1}\right\}$ has a perfect 1-k matching. Hence $G-\left\{x,y,y_1,\cdots ,y_{k-1}\right\}$ has a perfect 1-k matching.

Case 2 $xy\notin E\left(H\right)$

Let $X=\left\{x_1,\cdots ,x_m\right\}$ and $Y_i=N_H\left(x_i\right)=\left\{y_{i1},\cdots ,y_{ik}\right\},1\le i\le m$. Then $\left|Y_i\right|=k$. Let $G^{\text{*}}$ be the resulting graph obtained from G by shrinking every $Y_i$ to a single vertex $y_i^{\text{*}}$ and deleting the multiple edges. Then $G^{\text{*}}$ is a bipartite graph with bipartition $\left(X,\left\{y_1^{\text{*}},\cdots ,y_m^{\text{*}}\right\}\right)$. Then $M^{\text{*}}=\left\{x_1{y}_1^{\text{*}},\cdots ,x_m{y}_m^{\text{*}}\right\}$ is a perfect matching

of $G^{\text{*}}$. Since $\left|N_G\left(T\right)\right|>\frac{1}{k}\left|T\right|$ for any non-empty proper subset T of Y, we have that

$\left|N_{G^{\text{*}}}\left(T^{\text{*}}\right)\right|=\left|N_G\left({\displaystyle \underset{y_i^{\text{*}}\in T^{\text{*}}}{\cup }{Y}_i}\right)\right|>\frac{1}{k}\left|{\displaystyle \underset{y_i^{\text{*}}\in T^{\text{*}}}{\cup }{Y}_i}\right|=\left|T^{\text{*}}\right|$

for any non-empty proper subset $T^{\text{*}}$ of $Y^{\text{*}}$. Without loss of generality, suppose that $x=x_1$ and $y=y_{mk}\in Y_m$. By Proposition 1.1, $G^{\text{*}}-x_1-y_m^{\text{*}}$ has a perfect matching. Let $M_1^{\text{*}}=M^{\text{*}}-\left\{x_1{y}_1^{\text{*}},x_m{y}_m^{\text{*}}\right\}$. Then $M_1^{\text{*}}$ is not a maximum matching of $G^{\text{*}}-x_1-y_m^{\text{*}}$ and $x_m$, $y_1^{\text{*}}$ are $M_1^{\text{*}}$-unsaturated vertices. So there exists an $M_1^{\text{*}}$-augmenting path $P^{\text{*}}$ of $G^{\text{*}}-x_1-y_m^{\text{*}}$ joining $y_1^{\text{*}}$ and $x_m$. Without loss of generality, suppose that $P^{\text{*}}=y_1^{\text{*}}{x}_2{y}_2^{\text{*}}{x}_3{y}_3^{\text{*}},\cdots ,x_q{y}_q^{\text{*}}{x}_m$. Then we can assume that $P=y_{1k}{x}_2{y}_{2k}{x}_3{y}_{3k},\cdots ,x_q{y}_{qk}{x}_m$ is a path of $G-x_1-y_{mk}$ joining $y_{1k}$ and $x_m$. Let $M^{\prime }=\left(M-\left\{x_1{y}_{11},\cdots ,x_1{y}_{1k},x_m{y}_{mk}\right\}\right)\Delta E\left(P\right)$. Then $M^{\prime }$ is a perfect 1-k matching of $G-\left\{x,y,y_{11},y_{12},\cdots ,y_{1\left(k-1\right)}\right\}$.

This work is supported by the National Natural Science Foundation of China (No.12161073).