A question on graphs with all vertices are either central or peripheral If there are only two central vertices and rest are peripheral. Can we deduce that central vertices are always
adjacent? I got following examples (eccentricity mentioned) from where I proposed this. Is there any counter example for this?
Any hint or suggestion is hearty welcome. Thanks for the help. 

One more example in support :

 A: If I am not mistaken then this is a counter example:

The black vertices are central and non-adjacent. All other vertices are peripheral. Matching colors show that the corresponding vertices are of distance 3. The graphs diameter is also 3.
A: I found another solution which might be much more satisfying than just a counterexample like in my other answer. So I decided to write a new answer.
I found a way to construct a graph of diamater $n\geq 4$ and radius $n-1$ with exactly two central vertices. This is a counterexample to your question, as well as to your extended conjecture in the comment below my other answer, as this means we can make the distance between the central vertices as big as we want. The main idea is to start with a graph of diameter $n$ and then remove some vertices to lower the eccentricity of two chosen vertices.
Start with the edge-graph of the hypercube $Q_n$ (the graph consisting of nodes representing the binary sequences of length $n$, two nodes being adjacent if and only if their sequences differ in a single digit). $Q_n$ has $2^n$ vertices and is of diameter and radius $n$. Now, choose two non-antipodal vertices $v,w\in V(Q_n)$, i.e. $d(v,w)<n$. We can choose $d(v,w)=n-1$. These vertices will be our central vertices in the final graph. Note that in $Q_n$ any vertex $u$ has a unique antipodal vertex $u'$, i.e. a vertex of distance $n$. We construct our final graph $G$ by removing $v'$ and $w'$ from $Q_n$. Now, $G$ is of diameter $n$, radius $n-1$ with exactly two central vertices $v,w$ of distance $n-1$ to each other.
Here is what this looks like in the case of the 4-dimensional hypercube, i.e. $n=4$:

The graph is of diameter 4 and of radius 3. The black vertices are central and of distance 3 from each other. Matching colors indicate that the corresponding vertices are antipodal, hence define the diameter of the graph.
The restriction $n\ge 4$ is necessary, because otherwise it might happen that two formerly antipodal vertices are no longer joined by a path of length $n$. Note that this approach might also be generalized to $r$ central vertices, but then there has to be investigations on the lower bound of $n$ dependent on $r$.
