$G$ is a ${\Delta}$-regular graph with $n \le 2{\Delta}+1$, then $G$ does not contain a clique of cardinality greater than ${\Delta}$, since $2({\Delta}+1)>2{\Delta}+1\ge n$.

\[r:DisjointClique\] Let $G$ be any ${\Delta}$-regular graph with order $n$ and ${\Delta}\ge1$ such that $G$ has two disjoint cliques of cardinality ${\Delta}$. Then

1.  If $n=2{\Delta}$, then $G$ is isomorphic to the Cartesian product graph $P_2 \Box K_{\Delta}$.

2.  If $n=2{\Delta}+1$, then ${\Delta}$ is even (since $n{\Delta}=2m$) and $G$ can be obtained from $P_2 \Box K_{\Delta}$ by removing ${\Delta}/2$ copy edges of $P_2$ and connecting the ${\Delta}$ vertices with degree ${\Delta}-1$ with a new vertex. In particular, if $S$ is a clique of cardinality ${\Delta}$ in $G$, then $\overline{S}$ is not an exact defensive (${\Delta}-2$)-alliance.

\[t:An+D-2&lt;=n+m\] Let $G$ be any ${\Delta}$-regular graph with order $n$, size $m$, ${\Delta}\ge 3$ and $2{\Delta}\le n \le 2{\Delta}+1$. Then $n\le A_{{\Delta}-2}(G)\le n+m+2$.

Note that if ${\Delta}=3$ then $n=6$, and $G$ is a graph isomorphic to either $K_{3,3}$ or $P_2\Box K_3$. Thus, a simple computation gives $6\le A_{1}(K_{3,3})=15\le 6 + 9 + 2$ and $6\le A_{1}(P_2\Box K_3) = 11 \le 6 + 9 + 2$.

Assume now that ${\Delta}\ge 4$. Clearly, $G$ is a connected graph and $\operatorname{diam}G = 2$, since 