graph theory which will be useful.

\[Dirac\] A graph with order $n \ge 3$ is Hamiltonian if every vertex has degree $n / 2$ or greater.

In what follows we will use the following notation: for any $A,B\subset V$, we denote by $N(A,B)$ the number of edges with one endpoint in $A$ and the other endpoint in $B$.

\[t:An+D-2=n\] Let $G$ be any ${\Delta}$-regular graph with order $n < 2{\Delta}$. Then $A_{{\Delta}-2}(G)=n$.

Notice that ${\Delta}\ge 2$, since otherwise, such a graph $G$ does not exist; furthermore, $n\ge {\Delta}+1 \ge 3$. We have that $G$ is a Hamiltonian graph by Theorem \[Dirac\]. Besides, by Theorem \[t:PropRegular\] i), we have that $A_{{\Delta}-2}(G)$ is the number of connected induced subgraphs of $G$ with minimum degree ${\Delta}-1$. Let us consider $u\in V$ and define $S_u:=V\setminus\{u\}$. Since $G$ is a Hamiltonian graph, $\langle S_u\rangle$ is connected. Besides, we have ${\delta}_{S_u}(v)\ge {\Delta}-1 \ge {\delta}_{\overline{S_u}}(v) + {\Delta}-2$ for all $v\in S_u$ and the equality holds at $w\in N(u)$. So, $S_u$ is an exact defensive (${\Delta}-2$)-alliance in $G$ and $A_{{\Delta}-2}(G)\ge n$.

Seeking for a contradiction assume that there is an exact defensive (${\Delta}-2$)-alliance $S\subset V$ with $|S|\le n-2$. Notice that $|S|\ge{\Delta}>n/2$, by Theorem \[t:PropRegular\] i). Then, since any 