TITLE: A condition that might force biregularity of a bipartite graph
QUESTION [5 upvotes]: Let $\Gamma$ be a finite connected bipartite graph with colour classes $U$ and $V$ such that:

every vertex of $U$ has degree $n$, and $n\ge 3$;
every vertex of $V$ has degree at least $4$;
$\Gamma$ has diameter $4$;
for every vertex $u$ of $U$, there are at least $(n+1)/2$ vertices in the neighbourhood $\Gamma(u)$ of $u$ that have degree the minimum degree of $\Gamma(u)$.

Is $\Gamma$ biregular (i.e., the vertices of $V$ have constant degree)? If not, then under what natural and weak conditions is biregularity forced to hold?

REPLY [3 votes]: I tried to construct an example that is not biregular. To make it easy, I assumed that n=3 and that the vertices of V all have degree 4 except one of degree 5 (so each vertex of U is automatically adjacent to at least two vertices of degree 4).
One option where the numbers work out is to have 7 vertices in U and 5 vertices in V, so if I set V = {A,B,C,D,E} and then make the 7 neighbourhoods equal to {ABC, ABD, ABE, ACE, ADE, BCD, CDE} then I think that vertex A has degree 5, and all the others degree 4. 
I don't know what sort of condition you'd need to rule out examples like this..