$A_{{\Delta}}(G)=1$. Besides, if $A_{{\Delta}}(G)=1$, then there is an exact defensive ${\Delta}$-alliance $S$ in $G$ with ${\delta}_S(v) \ge {\delta}_{\bar{S}}(v) + {\Delta}\ge {\Delta}$ (i.e., ${\delta}_S(v)={\Delta}$ and ${\delta}_{\overline{S}}(v)=0$) for every $v\in S$. So, the connectivity of $G$ gives that $G$ is a ${\Delta}$-regular graph.

\[t:Regulars\] Let $G_1,G_2$ be two regular graphs. If $A(G_1;x)=A(G_2;x)$, then $G_1$ and $G_2$ have the same order, size, degree and number of connected components.

Let $n_1,n_2$ be the orders of $G_1$ and $G_2$, respectively, and ${\Delta}_1,{\Delta}_2$ the degrees of $G_1$ and $G_2$, respectively. Then, by Theorem \[t:PropRegular\] ii) and iii) we have $$n_1 - {\Delta}_1 = n_2 - {\Delta}_2 \quad \text{ and } \quad n_1+{\Delta}_1 = n_2+{\Delta}_2$$ and we conclude $$n_1 = n_2 \quad \text{ and } \quad {\Delta}_1 = {\Delta}_2.$$ Hence, both graphs have the same size. Finally, since $A_{{\Delta}_1}(G_1) = A_{{\Delta}_2}(G_2)$, they have the same number of connected components by Theorem \[p:AlliPoly\] vi).

\[c:Reg1\] Let $G_1,G_2$ be two regular graphs with orders $n_1$ and $n_2$, and degrees ${\Delta}_1$ and ${\Delta}_2$, respectively. If $n_1 \neq n_2$ or ${\Delta}_1 \neq {\Delta}_2$, then $A(G_1;x) \neq A(G_2;x)$.

The next theorem characterizes the degree of any regular graph by the number of non-zero coefficients of its alliance polynomial.

\[t:PolyReg\] Let $G$ be 