Are the graphs $C_{2n}^{n-1}$ strongly regular? 
Are the graphs $C_{2n}^{n-1}$ strongly regular?  If they are, are there any other strongly regular graphs with parameters $$(v,k, \lambda, \mu)=(v,v-2,v-4,v-2)\,?$$

Here $C_k$  is the cycle with $k$ vertices, and $G^k$ is the graph with vertex set $V(G)$, where any two vertices $u$ and $v$ are adjacent if the distance between $u$ and $v$ in $G$ is at most $k$. A regular graph $G$ of degree $k$ with $v$ vertices, is said to be strongly regular, denoted by $\text{srg}(v, k, \lambda, \mu)$, if any two adjacent vertices in $G$ have $\lambda$ common neighbors and any two non adjacent vertices in $G$ have $\mu$ common neighbors.
 A: It helps a lot if you look at what is $C_{2n}^{n-1}$ : In an even cycle any two vertices are at distance at most $n-1$, except for "opposite vertices" in the cycle which are at distance n.
We get that $C_{2n}^{n-1}$ is just the complete graph $K_{2n}$ minus a perfect matching. For example, for $n=4$, we get $C_8^3=K_8\setminus\{\text{red edges}\}$ :

From there it is easy to derive that it is indeed $srg(v,v-2,v_4,v-2)$

*

*it is $v-2$ regular,

*any two no adjacent vertices are opposite and have the same
neighbourhood, hence $\mu=v-2$,

*any two adjacent vertices do not have there respective opposite vertices as common neighbour so that $\lambda=v-4$.

Now for the second part of the question : no these are the only ones. it can be done by construction. Start with a set of $v=2n$ vertices $\{u_1,\ldots,u_v\}$ and no edges so far. Because $v_1$ has degree $v-2$ it is incident with all other vertices except one, call this one $v_{n+1}$. Because $\mu=v-2=d(v_{n+1})$, $v_1$ and $v_{n+1}$ are connected to all other vertices. Repeat for $(v_2,v_{n+2})$ and so on. We must end up with $C_{2n}^{n-1}$.
