induced bi-regular subgraph of a bi-regular bipartite graph I'm not very familiar with the classical theorems and results that are well known to experts in graph theory, hope you don't mind if my question is too naive.
Given bi-regular bipartite graph $G=(V,E)$, where $V=V_1 \bigcup V_2$ and $V_1 \bigcap V_2=\emptyset$. Suppose $\text{deg}(u)=m, \forall u\in V_1$ and $\text{deg}(v)=n, \forall v\in V_2$. It then holds that $m|V_1|=n|V_2|$.
The question is: suppose there exists an integer $k < m$ such that $k|V_1|=n L$ where $L < |V_2|$ is an integer, can we always find an induced bi-regular subgraph by carefully choosing $L$ vertices in $|V_2|$ and all vertices in $V_1$?
The above condition is actually the sufficient condition if there exists a bi-regular graph, with $k$ and $n$ being the degrees of vertices.
Thank you for your help!
 A: It is easy to deal with this problem in terms of covers of $V_1$. Namely, for each vertex $v_2\in V_2$  a set $N(v_2)$ of its neighbors is a subset of $V_1$ of size $n$. Each vertex $v_1\in V_1$ is covered by a family $\mathcal U=\{N(v):v_2\in V_2\}$ exactly $m$ times. Given $k<m$ such that $kn$ is divisible by $r=|V_1|$ we want to find a subfamily $\mathcal V$ of a given $n$-uniform cover $\mathcal U$, which covers each vertex of $V_1$ exactly $k$ times. This is not always possible. For instance, recently Andriy Yurchyshyn, solving an other problem, for $r\in\{6,10,14,18\}$ constructed a family $\mathcal U$ of  subsets of a set $[r]=\{1,\dots,r\}$ of size $n=r/2$ each, such that each element $v_1\in [r]$ is covered exactly $m=\tfrac 14 {r \choose r/2}$ times and  $\mathcal U$ contains no subfamily $\mathcal V$ which covers each element $v_1\in [r]$ exactly $k=1$ times. The family $\mathcal U$ for $r=6$ is  

124 235 346 451 562 613  123 345 561  246

The following subfamilies of $\mathcal U$ also produce  counterexamples:

124 235 346 451 562 613

and 

123 345 561 246

