is an exact defensive (${\Delta}-2$)-alliance in $G$, then $S$ is a clique of cardinality ${\Delta}$ and by Lemma \[l:clique\] there are at most $2$ exact defensive (${\Delta}-2$)-alliances with ${\Delta}$ vertices. Assume now that $n=2{\Delta}+1$. So, ${\Delta}$ is even. We only need to compute the possible exact defensive (${\Delta}-2$)-alliances in $G$ with cardinalities ${\Delta}$ and ${\Delta}+1$. If $S$ is an exact defensive (${\Delta}-2$)-alliance in $G$ with $|S|={\Delta}+1$, then ${\delta}_S(u)\ge {\Delta}-1$ for every $u\in S$ and ${\delta}_S(u_0)={\Delta}$ for some $u_0\in S$, since otherwise ${\delta}_S(u)={\Delta}-1$ for every $u\in S$ and we conclude $({\Delta}+1)({\Delta}-1)=|S|({\Delta}-1)=2m_S$, with $m_S$ the size of $\langle S\rangle$, which is not possible since ${\Delta}$ is even. Hence, $N(\overline{S},S)\le {\Delta}$; furthermore, since $|\overline{S}|={\Delta}$, ${\delta}_{S}(v)\ge 1$ for all $v\in \overline{S}$, and so, $\overline{S}$ is a clique. If $S$ is an exact defensive (${\Delta}-2$)-alliance in $G$ with $|S|={\Delta}$, then ${\delta}_S(u)\ge{\Delta}-1$ for every $u\in S$ and $S$ is a clique of cardinality ${\Delta}$. Lemma \[l:clique\] completes the proof since if $G$ has two cliques of cardinality ${\Delta}$, then they are disjoint and Remark \[r:DisjointClique\] gives that $\overline{S}$ is not an exact defensive (${\Delta}-2$)-alliance in $G$.

\[t:RegCharact\] Let $G$ be a ${\Delta}$-regular connected graph with order $n$ and let $G^*$ be a graph with order $n_1$ 