Diameter of a Johnson Graph I'm reading the book Algebraic Graph Theory, by Godsil and Royle, and I am currently doing the first chapter's exercises. In one of them I am tasked with determining the diameter of a Johnson Graph with $v > 2k$.
In the first chapter of the book they don't elaborate much on Johnson Graphs besides giving the definition. I don't know how to approach this. I've read on wikipedia that Johnson Graphs are $k(n-k)$-regular, and I know the number of vertices, could this be used to calculate the diameter? And if so, how would I go about showing its regularity to then calculate the diameter?
 A: The Johnson graph $J(v,k)$ is defined as follows.
The vertices of $J(v,k)$ are the $k$-element subsets of an $v$-element set $X$.
Two vertices $V,W\subset X$ are adjacent when $|V\cap W|=k-1$.
Proposition. If $v>2k$, then $\operatorname{diam}(J(v,k))=k$.
Proof.
Let $V=\{v_1,\ldots,v_k\}\subset X,W=\{w_1,\ldots,w_k\}\subset X$ be chosen such that $V\cap W=\varnothing$.
Let $V_i=\{w_1,\ldots,w_i,v_{i+1},\ldots,v_k\}$, $i=0,1,\ldots,k$.
We have $V_0=V$, $V_k=W$ and $V_i\cap V_{i+1}=\{w_1,\ldots,w_i,v_{i+2},\ldots,v_k\}$.
Hence vertices $V_i$ and $V_{i+1}$ are adjacent for $i=0,1,\ldots,k-1$.
This means that the distance between $V$ and $W$ is at most $k$.
Now let us prove that the distance between $V$ and $W$ cannot be less than $k$.
Let $U_0=V,U_1,\ldots,U_s=W$ be a path between $V$ and $W$ of length $s<k$.
Then
$$
|U_0\cup U_1|= k+1,
|U_0\cup U_1\cup U_2|\leq k+2,
\dots,
|U_0\cup U_1\cup\ldots\cup U_s|\leq k+s<2k.
$$
We get a contradiction since
$V\cup W\subset U_0\cup U_1\cup\ldots\cup U_s$ and
$|V\cup W|=2k$.
