Given an undirected graph $G = (V, E)$, a *vertex k-cut* of $G$ is a vertex subset of $V$ the removing of which disconnects the graph in at least $k$ connected components
(from <https://cris.unibo.it/handle/11585/713744>).


Given a biconnected graph, the minimal cardinality of a vertex 2-cut is 2, by definition.


My conjecture is that the above property can be generalised to:



> 
> The minimal cardinality of a vertex $k$-cut of a biconnected graph
> is $k$
> 
> 
> 


In other words, can one get $k$ connected components by removing less than $k$ vertices from a biconnected graph?



Minimal cardinality of a vertex $k$-cut of a biconnected graph
Question (if I got it right):



> 
> can one get $k$ connected components by removing less than $k$ vertices from a biconnected graph?
> 
> 
> 


Answer:


Yes. Consider the graph $G=(V,E)$ with $V=\{a,b,v\_1,...,v\_n\}$ and $E=\{av\_i,bv\_i : i=1,...,n\}$. It is 2-connected. Removing nodes $a$ and $b$ you get $n$ components, where $n$ can be arbitrarily large.


