Partitioning vertices in biregular graph This is the continuation of the previous question. Consider a $(a,b)$-biregular graph $(X,Y,E)$. In this case, $a|X|=b|Y|$ holds. Assume that $g:=|Y|/a=|X|/b$ is an integer. Further, assume that $|X|\geq|Y|$ (equivalently $b\geq a$). The problem is to find or to prove the existence a partition $\{X_i\}$ of $X$ with $|X_i|=g$ for all $i$ (so that $\bigcup_{i=1}^{b}X_i=X$) such that $|N(y)\cap X_i|=1$ for any $y\in Y$ and $i \in \{1,2,...,b\}$, where $N(y)$ is the neighborhood of $y$.
Intuitively, an example of this problem is the following: Assume that there are $C$ classes and $S$ students. Assume that $S \geq C$. Each student takes $c$ classes and each class has exactly $s$ students at a time. In this case, $sC=cS$ holds. Assume that there is an integer $g$ such that $g=C/c=S/s$. Can one group $g$ students into one group so that each class has only one student from each group?
 A: Here is a counterexample where the two sides are the same size, and neither can be partitioned.  (BTW thanks for these problems!  I had fun with both of them -- they are perfect for thinking while lying in bed.  :) )
All arithmetic modulo $6$.
$X = Y = \{0,1,2,3,4,5\}$, and $(x,y)\in E$ iff $y - x \in \{0, 1, 3\}$, i.e. neighborhood $N(x) = \{x, x+1, x+3\}$.  So this is a $(3,3)$-biregular graph, with $g = 6/3 = 2$.
We will show that $X$ cannot be partitioned.  The $Y$ case is similar.
In fact, we will show that any $x, x' \in X$ must share a neighbor in $Y$, i.e. not only the partition cannot exist, even one such size-$2$ subset (part) cannot exist. 
The neighborhoods of the various $x\in X$ are: 
$$\{0,1,3\}, \{1,2,4\}, \{2,3,5\}, \{3,4,0\}, \{4,5,1\}, \{5,0,2\}$$
It is easy to hand-check that each pair has a non-empty intersection.  Another way to see this is that, since each neighborhood is half of $Y$, two neighborhoods don't intersect iff they are complementary subsets of $Y$.  But the complement of $\{x, x+1, x+3\}$ is $\{x+2, x+4, x+5\}$, which is in the form $\{z, z+1, z+4\}$, so it isn't any of the six neighborhoods.  In other words, no two neighborhoods are complements of each other.
